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Thus glg;1 E AB and AB is a sub group of G. 4.63. Show that the intersection of two normal subgroups is a normal subgroup. Solution: We refer the reader to Problem 3.15, page 55, in which we proved that the intersection of two subgroups is a subgroup. If H, K are normal subgroups of G, then HnK is a subgroup of G. If cEHnK and gEG, then g-lcgEH as cEH and H<J G, and g-lcgEK as cEK and K <J G. Thus g-Icg E HnK, and so HnK is normal in G. e. Factor groups In Section 4.3a we mentioned that the concept of a coset sometimes gives rise to a new group. This occurs when, and only when, the group is normal. Let G be a group and N <J G. Let us denote by GIN (read as "G over N", or "G factor N", or "the factor group of G by N") the set of right co sets of N in G. We turn GIN into a groupoid by defining a binary operation as follows. Define a product of two co sets Na and Nb to be the coset Nab. This definition of multi plication depends on a and b. But it is conceivable that if Na = Nal and Nb = Nb l, that Nalb l # Nab. In such case, what would '.ve take for the "product" of the two cosets, Nalb l or Nab? What we must show is that the product of two co sets is uniquely defined by the formula NaNb = Nab when N <J G. If Nal = Na and Nb l = Nb, then al = na for some n E Nand bl = mb for some mEN. Accordingly, alb l = namb = n(ama-l)ab = lab l = n(ama- l). Since N <J G, ama- l EN and hence lEN. Thus albl E Nab. where Since the cosets form a partition, it follows that Nalb l = Nab. But this is just what we wanted to prove. Thus GIN with this binary operation is a groupoid. Is it an associative groupoid? ((Na)(Nb))(Nc) |
= (Nab)Nc N(ab)c = Na(bc) (as G is associative) = (Na)(Nbc) = (Na)((Nb)(Nc)) and so GIN is an associative groupoid. Theorem 4.17: GIN is a group. The mapping v: G ~ GIN defined by gv = Ng is a homo morphism of G onto GIN. Proof: First, GIN is an associative groupoid. N 1 = N is an identity, for Na' N1 = N(a '1) = Na and N· Na = N(1' a) = Na. Next, each element Na has an inverse, for NaNa- l = N(aa- l) = N while Na-lNa = N(a-la) = N. Thus GIN is a group. Clearly v is a mapping of G onto GIN. Since (glg2)v = N(glg2) and (glv)(g2v) = NglNg2 = Ng lg2, (glg2)v = (glv)(g2 v) and hence v is a homomorphism. Sec. 4.3] COSETS 115 v is called the natural homomorphism of G onto its factor group GIN. Note that in the case where G is a group whose binary operation is +, as we remarked at the end of Section 4.3a, the elements of GIN are of the form N + g. Instead of using. the multiplicative notation for GIN, we use additive notation. Our definition of the product of two cosets is written as the sum: (N + g1) + (N + g2) = N + (gl + g2). Problems 4.64. Prove that Z/E"", C2, where E is the set of even integers and C2 is the cyclic group of order 2. Solution: As Z is abelian, E is a normal subgroup of Z and Z/E makes sense. A coset of E is of the form E + z, z E Z. E + 0 consists of all the even integers, E + 1 of all the odd. Since an integer is either odd or even, these are all the co sets. gp({(E + 1)}) contains E + 1 and (E + 1) + (E + |
1) == E, and so gp({(E + 1)}) == Z/E. Thus Z/E is cyclic of order 2. Hence the result. 4.65. Is Q/Z "'" Q, where Q is the additive group of rationals? ments of Q/Z.) (Hint. Examine the order of the ele Solution: A coset of Z in Q is of the form Z + q, q E Q. As q == mIn for some integers m, n, it follows that nq E Z. Then (Z + q) + (Z + q) +... + (Z + q) == Z ~--------~----------~; n term~ in all and therefore Z + q is of finite order in Q/Z. Thus every element of Q/Z is of finite order. On the other hand no element other than 0 of Q is of finite order, for nq =1= 0 if q =1= 0 and n =1= O. It is this fact that we will utilize to prove that Q/Z is not isomorphic to Q. Suppose it were and that then there exists an element Z + r 0: Q/Z -4 Q was such an isomorphism. Choose q E Q (q =1= 0); of Q/Z such that (Z + r)o == q. Now Z + r is of finite order n, say. Then (n(Z + r»o == nq and now n(Z + r) == Z. Since Z is the identity of Q/Z and 0 is an isomorphism, 0 takes Z to the identity of Q, namely 0; hence nq == O. But q =1= 0, so nq =1= O. This contradiction proves that there exists no isomorphism of Q/Z onto Q. 4.66. Find S3/S~. Solution: From Problem 4.56, S~ == A 3. Hence the co sets of S3/S~ are A3 == A 3, == {" 0'1> O'2} A37"l == h1' 7"2' 7"3}' Thus S3/S~ consists of these two cosets. The multiplication table is and A3 A37"1 4.67. Prove that if G "'" H, then G' "'" H' and Z(G) "'" Z(H |
). Solution: Let 0 be an isomorphism from G to H. Then 'II == 0IG' is a monomorphism into H, since 0 is both one-to-one and a homomorphism. What is G'¥? The elements of G' are products of commutators and their inverses. As the inverse of a commutator is a commutator, each element of G' is of the form Cl ••• Ck, where each Ci is a commutator. Then the images of G' under'll are of the form Cl¥ •.• Ck¥' If C == [a, b] == a- 1b- 1ab, then C¥ == (ao)-1(bo)-1aobo == lao, bo] Thus the Ci¥ are commutators and so G'¥ C H'. 116 ISOMORPHISM THEOREMS [CHAP. 4 To show G'>Jr = H', we need only show that all possible commutators [ht> h21 are images under 'iF. As 0 is onto, there exists gl' g2 such that glO = hI' g20 = h2. Hence [gt> g21>Jr = [ht> h21. There fore G'>Jr = H' and consequently G' == H'. Let ¢ = 0IZ(G)' Then ¢ is a monomorphism of Z(G) into H, and we need only show that it maps onto Z(H) to prove Z(G) == Z(H). Let z E Z(G). For each hE H there exists g E G with go = h. Hence zoh = zogo = (zg)o = (gz)o = gozo-= hzo and so Z(G)¢CZ(H). Let x E Z(H). As 0 is onto, there exists y E G with yo = x. Let g E G. Then (gy)o = goyo = yogo = (yg)o. As 0 is one-to one, gy = yg. Hence y E Z(G), and every element of Z(H) is an image of an element of Z(G). The result follows. 4.68. If A is abelian show that A/A' == A. (i) (ii) Show that G/G' is |
abelian. (iii) Show that G/N abelian implies N -:2 G'. Solution: (i) A' is the subgroup generated by the commutators x-Iy-Ixy, x,y E A. As A is abelian, x-Iy-Ixy = y-Ix-Ixy = y- 11y = 1 and A' is the subgroup generated by 1. Since any prod uct of 1 and its inverse is again 1, A' = {1}. Let p be the natural homomorphism of A to A/A'. To show P is an isomorphism we need only show that it is one-to-one. Suppose alP = a2P, i.e. {1}al = {1}a2, i.e. {al} = {a2}. Then of course al = a2' so p is one-to-one. Thus P is an isomorphism. (ii) Let G'x and G'y be two elements of G/G'. Then G'xG'y = G'xy while G'yG'x = G'yx. Now (x-I)-I(y-I)-lx-1y-1 = xyx-1y-1 E G', and so G'yx contains the element xyx-Iy-l. yx = xy. (G'x)(G'y) = (G'y)(G'x) and G/G' is But G'xy contains xy; hence G'xy = G'yx. Therefore abelian. (iii) We need only show that N contains every commutator. For then N contains the subgroup generated by the commutators, which is G' by definition. If G/N is abelian and x, yare any elements of G, (xy)N = xNyN = yNxN = (yx)N. Hence xy = yxn where n E N. Multiplying on the left by y-l and then by x-I, we obtain x-1y-Ixy = n, i.e. [x,yl EN. 4.69. Show that if H is a subgroup of G containing G', then H <J G. Show that if H is of index 2 in G, then H |
<J G and G/H is cyclic of order 2. Solution: Let hE H and consider g-lhg. Now g-lhgh- l is a commutator, and hence belongs to G' and thus to H. Therefore g-lhgh- l • h = g-lhg E Hand H <J G. If now H is of index 2, G = HUHg where HnHg = 0. Let hE H. If kEG, then k = hI or k = hlg (hI E H). Hence or where h2 E H. If g-lh2g E H, we are through. Otherwise g-lh2Y E Hg, so that g-lh2fl = h'g and thus g-lh2 = h'. Hence g = h2h- 1 E H. But this contradicts the assumption (h' E H) HnHg = 0. Thus we are forced to conclude that k-1hk E H for every hE H and every kEG, i.e. that H <J G. The two cosets of H in G are Hand Hg for some g Iii!: H. Accordingly gp(Hg) = G/H and so G/H is cyclic of order 2. 4.70. Show by considering a suitable non-normal subgroup H of S3 that the product of two co sets cannot be defined as on page 114 without ambiguity. Solution: We use the notation of Section 3.3a, page 57, in dealing with the symmetric group S3 of degree 3. Let us take H = gp(71) = {1,71}' HUI = {ut> 73} = H73 while HU2 = {U2,72} = H72' In page 114 we defined the product of the coset HUI and HU2 as HUIU2 = H. However, the product of H73 and H72 would be H7372 = HUI 0;6 H. This means that the product of cosets is not uniquely defined. Sec. 4.4] l'IOMOMORPHISM THEOREMS 117 4.4 HOMOMORPHISM THEOREMS a. Homomorphisms and factor groups: The homomorphism theorem We now consider the connection between homomorphisms and factor groups. We have already |
established in Theorem 4.17 that corresponding to every factor group GIN there is a homomorphism v: G -? GIN such that Gv = GIN. What about the converse? Suppose now that 0 is a homomorphism of G into a group H. We ask: is there a normal subgroup N of G such that GIN ==' GO? Let us define the kernel of 0 (denoted KerO) by KerO = {g I g E G, gO = I}. We will show that KerO will do the trick. Theorem 4.18 (Homomorphism Theorem, also called the First Isomorphism Theorem): is a homomorphism of a group G into a group H, then is a normal subgroup of G, and 'r]: gO -? Ng defines an isomor If 0: G -? H N = KerO phism of Go onto GIN. Proof: First we will show that N is a subgroup. If gl, g2 EN, then (glg;:I)O = (gIO)(g;:IO) = (gIO)(g20)-1 = 1·1 = 1 and so glg;:l EN. Also 1 EN, so N oF~, the empty set. Thus N is a subgroup of G. To prove that it is a normal subgroup of G, let n EN and g E G. We must show that rIng EN; this will hold if (g-lng)O = 1. But (g-lng)O = (gO)-I(nO)(gO) = (gO)-II(gO) = (gO)-I(gO) = 1 Hence g-lng EN and N is a normal subgroup of G. Next we must show that 'r]: gO -? Ng defines a mapping. It is conceivable that there exist gl oF g2 with glO = g20. We ask: is Ng I = Ng2? For if not, 'r] is not a mapping as it is not uniquely defined. Now glg;:l EN, for (glg;:l)(J = (gIO)(g20)-1 = 1 as glO = g20. Hence glg;:l EN and gl = ng2 where n EN. Thus Ng I and Ng2 have gl in common and, as the |
right co sets form a partition, N gl = N g2. Consequently 'r] is a well defined mapping. Is it a homomorphism? (gIOg20)ry = ((glg2)O)ry = N(glg2) = Ng INg2 = (gIO)ry(g20)ry and so 'r] is a homomorphism. Finally, is 'r] one-to-one? If (gIO)ry = (g20)'r], then N gl = N g2. Then ngl = g2 for some is one-to-one and hence is an n E N, and g20 = (ngl)(J = nOglO = 1· glO = glO. Thus 'r] isomorphism. Problems 4.71. Let fJ be the homomorphism of Z into the mUltiplicative group of nonzero rational numbers defined by XfJ = 1 if x is even, and XfJ = -1 if x is odd. Find the kernel of fJ and examine the claim G/(Ker fJ) == GfJ. Solution: Ker fJ = {x I xfJ = I} = {x I x is even} and G/(Ker fJ) = {Ker fJ, Ker fJ + I}, so Ker fJ + 1 (Ker fJ + 1) + (Ker fJ + 1) = Ker fJ; hence Ker fJ + 1 is of order 2 generates G/(Ker fJ). We have and G/(Ker fJ) is cyclic of order 2. Now GfJ = {I, -I}. GfJ = gp({-l}) and, since (-1)' (-1) = 1, GfJ is also cyclic of order 2. Therefore G/(Ker fJ) and GfJ are isomorphic by Theorem 4.7, page 103. 4.72. Check that the kernel of the natural homomorphism of the additive group of integers (Z, +) onto Z/2Z is 2Z, where 2Z = {x I x = 2z, z E Z}, Solution: i.e. 2Z is the set of even integers. The natural homomorphism p is defined by ZP = 2Z + z. The identity of Z/2Z is the coset 2Z |
. if and only if 2Z + z = 2Z and hence if and only if z E 2Z. Thus Consequently z E Ker p Kerp = 2Z. 118 ISOMORPHISM THEOREMS [CHAP. 4 4.73. Verify that if G is any group, the subgroup generated by the squares of the elements of G, i.e. is a normal subgroup of G. elements of the form yy = y2, Solution: Let S denote the subgroup generated by the squares of the elements of G. Let xES. Then x is a product 8182 ••• 8k of elements of G each of which is a square or the inverse of a square. Since the inverse of a square is also a square, we may assume each 81> •••, 8 k is a square. If y E G, then y-1xy = y-181yy-182Y'" y-18ky = t 1t2 •• ·tk where ti = y- 18 iY. We assert that each ti is a square. Since 8i = rf for some ri> and hence the ti are squares as asserted and y-1xy E S. Thus S <l G. ti = y-18iY = y-1riyy-1riY = (y-lriy)2 4.74. Let 0' be the homomorphism of Q*, the multiplicative group of nonzero rationals into Q* defined by xu = Ixl. Find the kernel and image of 0'. Verify the homomorphism theorem directly in this case. Solution: 0' is a homomorphism, since (xY)O' = Ixyl = Ixllyl = (xO')(yO'). KerO' = {x I xu = l} = {x I Ixl = l} = {l,-l}. Q*O' = {x I x = yO', Y E Q*} = {x I x E Q*, x is positive} Let p be the mapping of Q*/(Ker 0') into Q*O' that takes (KerO')q ~ qO' Now (Ker O')q = {q, -q}, and so the only other "representation" for the coset (Ker O')q is (Ker O')(-q). Hence the other possibility for p as far as ( |
Ker O')q is concerned is that «Ker O')q)p = (-q)O'. But (-q)O' = lql = qO'. Thus p is a well defined mapping of G/Ker 0' into Q*O'. Since Iqlq21 = Iqlllq21 [(Ker O')ql • (Ker 0')q2]p = «Ker 0')qlq2)p = p is a homomorphism. Is p one-to-one? If (Ker O')qlP = (Ker 0')q2v, then we have that qlO' = q20" i.e. Iqll = Iq21· Thus if ql =1= q2' ql = -q2' Therefore (Ker O')ql = (Ker 0')q2' Hence p is one-to-one and it follows that Q* /(Ker 0') "" Q*O'. = «Ker O')ql)p· «Ker 0')q2)P 4.75. Let G be the group of mappings of the real line R onto itself of the form aa, b: x ~ ax + b, a =1= 0, a, b real numbers, x E R. Prove that the map is a homomorphism of G into G. Find the kernel and the image of this homomorphism and exhibit the isomorphism described in the homomorphism theorem. fJ: aa,b -> aa,O Solution: Note that aa,bac,d - since xaa,b = ax + band (xaa,b)aC,d = c(ax + b) + d. Then (aa,bac,d)fJ = (aac, bc+d)fJ = aac,O' (aa,bfJ)(ac,d fJ ) = aa,oac,o = aac,O' and so fJ is a homomorphism. If aa,b E Ker 9, we have aa,b9 = al,O as a LO = I, the identity mapping. Hence {aLb I b any real number} = Ker fJ. The image of aac, bc+d' A typical coset is fJ = {aa,o I a any nonzero real number}. |
(Ker fJ)ac,d = {a1,b I b any real number}ac,d {a c, bc+d I b any real number} {ac,e I e any real number} The isomorphism '1J between G/(Ker 9) and G9 as given by the homomorphism theorem is the one that takes the coset (Ker 9)ac,d to ac,o' 4.76. Prove that if G is cyclic of order nand p divides n, then there is a homomorphism of G onto a cyclic group of order p. What is the kernel of this homomorphism? Solution: Let G = yp(x) and let n = pm. Let H be cyclic of order p, H = yp(y). We define 8 to be the mapping Sec. 4.4] HOMOMORPHISM THEOREMS 119 (J is well defined as we established in Lemma 4.5 that the elements xi, 0 "" i "" n - 1, are all the dis tinct elements of G. Now let i and j be less than n. We have (xixi)(J = (xi+i-En)(J where i + j "" n - 1 while € = 1 if i + j :=:, n. Then € = 0 if (xix;)(J = yi+i-m = yiyiy-m Since the order of y divides n, y-en = 1. Hence (xixj)(J = yiyi = (xi(J)(xj(J) and so (J is a homomor phism. The kernel of (J is the set of all Xi such that Xi(J = 1, 0 "" i "" n - 1. Since xif) = yi and yi = 1 if and only if p divides i, Ker (J = {xi I p divides i} = {xp, x 2p,..., x(m-l)p} = gp(xP) 4.77. Let (R+, 0) be the multiplicative group of positive real numbers. Prove that the mapping of R+ into (R, +), the additive group of real numbers defined by f): x -> loglo x, is a homomorphism. What is the kernel of theorem, prove that (R+,o) == (R, +). |
Solution: f)? What is the image? Using the homomorphism (J is certainly a mapping of R+ into R. Furthermore, (xy)(J = loglo (xy) loglo X + loglo Y = X(J + y(J and so (J is a homomorphism. Ker(J = {x I loglox = 0, x E R+} = {1} We assert that R + (J = R. To see this observe that if x is any real number, then Moreover, log lOx = x. Then if y is any element of R, 10 Y(J = log lOY = y and hence R+(J = R. lOx E R +. The homomorphism theorem states that R+/(Ker(J) == R+(J = R. As Kerf) = {1}, all we must do is show that R+ /{1} == R+. We can indeed show this in general: if G is any group, G/{l} == G. To do this we exhibit the isomorphism. Let p be the natural homomorphism of G onto G/{l}. Then i.e. {l}gl = {l}gz. Then {gl} = {gz} we need only show that p is one-to-one. Suppose glP = gzP, and consequently gl = g2' Hence P isomorphism. Accordingly R+ == R+/{l} == R and so R+ == R. is one-to-one and thus an 4.78. Prove that the mapping (J: x -> eX defines an isomorphism of (R, +) onto (R+, 0), the multiplicative group of positive real numbers. Solution: (x + y)(J = eX+ Y = eXeY = (x(J)(y(J) and so (J is a homomorphism. If x(J = y(J, then eX = eY and eX - Y = 1, from which x - y = 0 and x = y, so (J is one-to-one. Is (J onto? Yes, for if y is any positive real number, the equation eX = y has a solution x E R. Thus (J is an isomorphism between (R, +) and (R+, 0). 4.79. Prove that [fg |
, a] = g-l[f, a] gig, a] for any f, g and a in a given group. Suppose G'r;;;,Z(G), center of G. Let a be a fixed element of G. Prove that the mapping (J: g -> [g, a] (Difficult.) phism of G into G. What is Ker (J? the is a homomor Solution: Observe that [fg, a] = g-lf-1a-1fga. On the other hand, g-l[f, a] g[g, a] g-1(f- 1a- 1fa)g(g-la- l ga) g-l f-1a-1f(agg-1a -l)ga g-lf-1a-1fga = [fg, a] If G' r;;;, Z(G), [f,aj[g,a]. Hence (fg)(J = f(Jg(J and (J is a homomorphism of G into G. Ker(J = {g I [g,a] = 1, g E G}. Thus Ker (J = C(gp(a», the centralizer in G of gp(a). [f, a] E Z(G) and r l [ f, a]g = [f, a]. Therefore [I, a]g-lg = [fg, a] then 4.80. Prove that if (J: G -> K is a homomorphism and IGI < 00, then IG(JI divides IGI. Solution: By the homomorphism theorem, G/(Ker (J) == G(J. By Lagrange's theorem, IKer (JI divides the order of G. Therefore IG(JI = IKer (JI divides IGI· IGI 120 ISOMORPHISM THEOREMS [CHAP. 4 4.81. Show that the group M of Mobius transformations is a homomorphic image of the group 'U = {(: ~) I a, b, c, d complex numbers, ad - bc = 1} Find the kernel of the homomorphism. Solution: In Problem 3.48, page 80, we showed M = {u(a, b, c, d) I a, b, c, d complex numbers |
, ad - bc = 1} We define the mapping f.L: 'U ~ M by (: ~) f.L = u(a, b, c, d). Clearly f.L is a mapping. Further more, using the results of Problem 3.46, page 79, = U(a1a2 + C1b2' b1a2 + d1b2, a1c2 + C1d2, b1C2 + d1d2) U(a1' b1, C1' d1)u(a2' b2, c2, d2) = (:~ ~~) f.L ( : : ~:) f.L Thusf.Lisahomomorphism. Kernelf.L = {G ~), (-~ _~)}, because (: ~)f.L = u(a,b,c,d) (the identity of M) if and only if a = d = 1 and b = c = ° or and u(a, b, c, d) = u(l, 0, 0, 1) a = d = -1 and b = c = 0, by Problem 3.49, page 80. Therefore M ~ 'U / {G ~), (-~ _~)} h. Correspondence Theorem. Factor of a factor theorem Let 0: G ~ K be a homomorphism of G onto K. If H is a subgroup of G, then HO is a subgroup of K. If H is normal in G, HO is normal in K. (See Problem 4.82 below.) What about the reverse of this procedure? Would the preimage of a subgroup S of K, i.e. the set {g I g E G, gO E S}, be a subgroup of G? We know that if S = {I}, then the preimage of S is Ker 0, and this is indeed a subgroup of G. We generalize this result in the following theorem. Theorem 4.19 (Correspondence Theorem): Let 0: G ~ K be a homomorphism of G onto K. The preimage H of any subgroup S of K is a subgroup of G containing Ker O. If S <J K, then H <J G. Furthermore if H1 is any other subgroup of G containing Ker 0 such that H 10 = S, then H |
1 = H. Proof: H = {g I gO E S}. Since l(} is the identity of K, and S contains the identity of K, then 1 E H and so H =1= p, the empty set. Also, if g, hE H, (gh- 1)O = gO(hO)-I. As gO E Sand ho E S, it follows that (gh -1)0 E S. Hence gh -1 E Hand H is a subgroup of G. Since KerO = {x I xO = I} and 1 E S, KerO CH. If S <J K, we must show that H <J G. Let hE H, g E G; then (g-lhg)O = (gO)-l(hO)(gO). Now hO E S, gO E K, and S <J K implies (g-lhg)O E S. Then g-lhg E H and so H <J G. Let H1 be a subgroup of G containing Ker 0 and suppose H 10 = S. We will show that H1 = H. Let h1 E H 1; then h 10 E S. Now H = {g I gO E S}. Therefore h1 E Hand hence HI C H. On the other hand if hE H, then hO = 8 E S. Choose hI E HI such that k l O = 8. Then hh;1 E Ker 0 C HI and so hE HI and H C HI. Thus H = HI. Sec. 4.4] HOMOMORPHISM THEOREMS 121 Corollary 4.20: Let 0: G ~ K be an onto homomorphism. Let S be a subgroup of K of index n < <Xl. Let H be the preimage of S. Then H is 6f index n in G. Proof: Let SkI, Sk2, •••, Skn, where ki E K, be the distinct co sets of S in K. As 0 is an onto homomorphism, there are elements gl,..., gn of G such that giO = ki. We claim that (4.5) are the distinct cosets of H in G. Suppose that Hgi = Hg j; then gigJ- I E H. Hence giO(gjO)-1 E |
S, i.e. kik j- l E S, from which Ski = Skj and i = j. Thus we have shown that all the co sets in (4.5) are distinct. Let g E G. Then gO E K and so gO E Ski for some integer i. Hence gO = Ski with l = s. Consequently ggi l is in the pre s E S. Consider x = ggil. xO = gO(giO)-1 = skiki image of S, so that ggi- l E H. This means that g E Hgi. We have thus shown that every element of G is a member of one of the co sets in (4.5). It follows that (4.5) consists of all the co sets of H in G. Of course we can always reformulate results about homomorphisms with the aid of the homomorphism theorem as results about factor groups. Thus we have Corollary 4.21: Let N <J G. Let L be a subgroup of GIN. Then we can write L = HIN where H is a subgroup of G containing N. If L <J GIN, then H <J G. If Ht/N = HIN where HI and H are subgroups of G containing N, then HI=H. Proof: Let v be the natural homomorphism of G ~ GIN. Let H = {g I gv E L}. Then by the correspondence theorem H is a subgroup of G; and if L <J GIN, H <J G. Also, H v = L. Since v: g ~ N g, H v consists of all cosets Nh, h E H. Because H::> N = Ker v by the correspondence theorem above, HIN makes sense and consists of all the cosets Nh, h E H. Hence HIN = Hv = L. Now if HI::> Nand Ht/N = HIN, then Hlv = L. It then follows by the correspondence theorem that HI = H. (Problems 4.82-4.89 below may be studied before reading Theorem 4.22.) The reader may very well wonder what happens when we take a factor group of a factor group. For example, if N <J G and GIN is a group containing a normal subgroup MIN, then what is (GIN)/(MIN)? The next theorem tells |
us that this is isomorphic to a single factor group, i.e. a factor G by one of its normal subgroups. Theorem 4.22 (Factor of a Factor Theorem, also called the Third Isomorphism Theorem): If in the factor group GIN there is a normal subgroup MIN, M::> N, then M <J G and GIM "'" (GIN)/(MIN) Proof: Let v: G ~ GIN be the natural homomorphism of G ~ GIN. Let p: GIN ~ (GIN)/(MIN) be the natural homomorphism of GIN ~ (GIN)/(MIN). Put 0 = vp. Then 8 is a homomorphism of G ~ (GIN)/(MIN); and since v is onto GIN and p is onto (GIN)/(MIN), vp is onto (GIN)/(MIN). Therefore GIKer (vp) "'" (GIN)/(MIN), by the homomorphism theorem. If g E G, gv = Ng and (Ng)p = (MIN)(Ng); note that here (MIN)(Ng) is a coset of the normal subgroup MIN in (GIN), i.e. an element in the group (GIN)/(MIN). Now the elements of MIN are all the cosets Nm, m EM. The identity of (GIN)/(MIN) is (MIN). We ask, what is the kernel of vp? It will be all g E G such that Uvp = (MIN)Ng = MIN. But in 122 ISOMORPHISM THEOREMS [CHAP. 4 that case Ng E MIN, i.e. Ng = Nm for some mE M. Hence g = nm where n EN. But M d N. Therefore gEM and so Kervp eM. Note that if mE M, m(vp) = (MIN)Nm = MIN. Then Ker vp = M. Thus M as a kernel of a homomorphism is normal in G and GIM ==0 (GIN)/(MIN), which is the required result. Problems 4.82. Prove that if (J: G -> K subgroup of K. If H <l G, prove that H(J <l K. is a homomorphism of G onto K, and H is a subgroup of G, then H |
(J is a Solution: Since 1(J E H(J, H(J # 0. If Xl' X2 E H(J, Xl = hl(J, X2 = h2(J for some hI' h2 E H. Then where h = hlh;:l E H. Hence H(J is a subgroup of K. If now H <l G, for all g E G and all hE H. Now any element of K, say k, is of the form g(J for some g E G; and any element of H(J is of the form h(J. Is (g(J)-lh(Jg(J E H(J? Yes, because (g(J)-I(h(J)g(J = (g-lhg)(J and as g-lhg E H, (g(J)-I(h(J)g(J E H(J. Thus H(J <l H. then g-lhg E H 4.83. Let G = D4 = {O"l' 0"2, 0"3, 0"4, 1', 1'0"2' 1'0"3' T0"4}' Let K = gp(b) be cyclic of order 2. Then (J: G -> K defined by O"i(J = 1, (TO"i)(J = b, i = 1,2,3,4, is a homomorphism of G onto K. (Take this as a fact.) Find all the subgroups of K and all their preimages. Check that the assertions of Theorem 4.19 and Corollary 4.20 hold. (See Problem 3.42, page 77, for the multiplication table.) Solution: The subgroups of K are KI = K and K2 = {1}. GI = the preimage of KI = {g I g(J E K} = G. G2 = the preimage of K2 = {g I g(J = 1} = {O"lo 0"2, 0"3' 0"4}' (a) GI, G2 are subgroups of G containing Ker (J = G2• (b) KI and K2 are normal in K, and GI and G2 are normal in G. (c) The sub |
groups of G containing Ker (J are GI and G2. We note then that Gi(J = Gj(J implies i = j where 1 "" i, j "" 2. (d) KI is of index 1 in K. GI is of index 1 in G. K2 is of index 2 in K. G2 is of index 2 in G, its co sets being GI = {O"lo 0"2' 0"3' 0"4} and TG 2 = {T, 1'0"2' 1'0"3' T0"4}' Thus (a), (b), (c) and (d) agree with Theorem 4.19 and Corollary 4.20. 4.84. Let G = gp(a) be cyclic of order 12 and let K = gp(b) be cyclic of order 4. Let (J: G -> K be defined by ai(J = bi, i = 0,1,2,...,11. Then (J is a homomorphism of G onto K. (Take this as fact.) Find all the subgroups of K and all their pre images. Check that the assertions of Theorem 4.19 and Corollary 4.20 hold. Solution: The subgroups of K are Kl = K, K2 = {1, b2}, K3 = {1}. GI = the preimage of KI = x(J = b2}. Clearly 1 E G2• {x I X(J E K} = G. G2 = the preimage of K2 = {x I X(J = 1 or a(J = b # b2, hence a (;l G2• a2(J = b2, so a2 E G2. Continuing in this fashion we conclude that G2 = {1, a2, a4, a6, as, aIO}. Finally, G3 = the preimage of K3 = {x I x(J = 1} = {1, a4, as}. (a) Glo G2 and G3 are all subgroups of G containing Ker (J = Gg • (b) K l, K2 and Kg are normal in K, and GI, G2 and G3 are normal in G (trivially, as G is abelian). (c) The subgroups of G containing the kernel of (J |
are those containing a4 • Hence the subgroups of G containing Ker (J are GI, G2 and Gg. Note Gi(J = Gj(J implies i = j (1 "" i, j "" 3). (d) Kl is of index 1 in K. GI is of index 1 in G. K2 is of index 2 in K, its co sets being K2 = {1, b2} and K 2b = {b, bg}. G2 is of index 2 in G, its cosets being G2 = {1, a2, a4, a6, as, aIO}, and Gza = {a, a3, as, a7,a9, all}. Kg is of index 4 in K. Gg is of index 4 in G, the distinct cosets being Gg = {1, a4, as}, Gga = {a, as, a 9}, Gga2 = {a2, a6, alO}, Gga 3 = {ag, a7, all}. Thus (a); (b), (c) and (d) agree with Theorem 4.19 and Corollary 4.20. Sec. 4.4] HOMOMORPHISM THEOREMS 123 4.85. Let 0: G --> K be an onto homomorphism. Let K be cyclic of order 10. Prove that G has normal subgroups of index 2 and 5 and 10. Solution: Let K = gp(k). Then the subgroups KI = {1}, K z = gp(k5), Ka = gp(k2) are normal subgroups of K of index 10, 5 and 2 respectively. Consequently their preimages, by Theorem 4.19 and Corollary 4.20, are normal of index 10, 5 and 2 respectively. Hence the result. 4.86. Let N <l G and suppose GIN is cyclic of order 6. Let GIN = gp(Nx). Find all the subgroups of GIN and express them in the form of Corollary 4.21. (Hard.) Solution: Let GIN = K. Let K j = {N}, K2 = {N, N x3}, Ka = {N, Nx2, NX4} and K4 = K. These are all the subgroups of K. To find the corresponding subgroups of Corollary 4. |
21 we let P: G --> GIN i = 1,2,3,4. be the natural homomorphism, i.e. gP = N g. Then let Gi be the preimage of K;, G j G2 Ga G4 {g I gP = N} = {g I Ng = N} = {g I g E N} = N {g I gP = N or g = Nxa} = {g I Ng = N or Ng = Nxa} = NuNxa {g I gP = N or gP = Nx2 or gP = NX4} = NuNx 2uNx4 Then GJN = Ki for i = 1,2,3,4. {g I gP E K} = G. 4.87. Use the correspondence theorem to prove that if H is a subgroup of G containing G' (the derived group of G), then H <l G (i.e. prove Problem 4.69 by another method). Solution: Let P: G --> GIG' be the natural homomorphism; then P is onto. GIG' is abelian by Problem 4.68. Any subgroup of GIG' is therefore normal, and thus Hp = S, say, is normal in GIG'. By the correspondence theorem, H is the preimage of S. Hence using the correspondence theorem once more, H is normal in G. 4.88. 4.89. Let H be a subgroup of index n in G. Let 0: G --> K be a homomorphism onto K. Prove that Ho is of index n in K if H d Ker o. Solution: It is only necessary to prove that S = Ho is of finite index in K, for then the result follows from Corollary 4.20. If Hg lo..., Hgn are the co sets of H in G, then we claim that {S(gjo),..., S(gno)} is the set of all the co sets of S in K. We need only show that if k E K, then k E S(giO) for some i = 1,...,n. As 0 is onto, there is agE G such that go = k. Let g = kg i• Then k = go = kO(giO) E S(giO). The result now follows from Corollary 4. |
20. Alternatively we can show that S(gIO),..., S(gno) are all distinct. Suppose S(giO) = S(gjo). Then (gio)(gl)-l = (g;llj-I)O E S. Hence gig;l E H as H, by the correspondence theorem, is the preimage of S. Accordingly i = j and the index of S in K is n. Let G be a group and let N be a normal subgroup of G. Suppose further that Land M are sub groups of GIN. Then show that we can write L in the form HIN, and M in the form KIN, where Hand K are subgroups of G containing N. Show also that if L (;; M, H (;; K; and if L <l M, H <l K. Show that if L (;; M and [M: L] = n < co, then [K: H] = n. Solution: This is just an application of Corollary 4.21 to Corollary 4.20 and Theorem 4.19. That L = HIN and M = KIN follows from Corollary 4.21. If p is the natural homomorphism, we recall that H = {g I gP E L} and K = {g I gP EM}. Hence if L (;; M, H (;; K follows immediately. Now if L <l M, we consider the homomorphism 0: K --> KIN defined by ko = kp i.e. 0= PIK' Clearly Ko = KIN and the preimage of L is H, the pre image of M is K. We can then conclude from the correspondence theorem that H <l K. for all k E K, 4.90. Let G = D 4 • Let M = {Ulo Ua, TUa, d, N = {UI' ua}. Then accept N <l G and M <l G. Consider GIN and MIN. Find (GIN)/(MIN) explicitly and check that it is isomorphic to GIM. In other words, check agreement with the factor of a factor theorem. (Use table on page 77.) Solution: GIN consists of the cosets Al = N = {Ulo ua}, A2 = U2N = {U2, U4}' Aa = TN = {T, TUa}, A |
4 = (TU2)N = {TU2,TU4}' Now M = {UI' Ua, TUa,T}, hence MIN consists of the elements AloA3' 124 ISOMORPHISM THEOREMS [CHAP. 4 MIN <l GIN. Therefore we can talk of (GIN)/(MIN). The elements of this group are the cosets of (MIN) in (GIN). These cosets are B! = (MIN)A! = {Av A 3} and B2 = (MIN)A2 = {A!AZ' A 3A z} = {A 2, A 4 }. plication in the group (GIN)/(MIN) is calculated in the usual way for cosets, e.g., B2B2 = (MIN)A 2(MIN)A 2 = (MIN)(A 2A 2) = (MIN)A! = B! (A3A2' for example, is calculated as follows: A3A2 = (rN)(u2N) = (ru2)N = A 4.) Multi as A2A2 = (U2N)(U2N) = U3N = N = A!. group of order 2 generated by B 2 • It is clear then that (GIM)/(MIN) = {B!,B 2} is the cyclic Now let us decide what GIM is. The elements of GIM are the co sets C! = M = {u!, U3' rU3' r} is the cyclic and C2 = MU2 = {U2, U4, rU4, rU2}' As C2C2 = MU2Mu2 = MU3 = M, GIM = {C v C2} group of order 2. Therefore GIM "'" (GIN)/(MIN). 4.91. Let G be the cyclic group of order 12, say G = gp(a). Let M = gp(a2), N = gp(a6). Consider GIN and MIN. Find (GIN)/(MIN) explicitly and check that it is isomorphic to GIM. In other words, check agreement with the factor of a factor theorem. (Difficult.) Solution: GIN consists of the co sets A! = N = {1, a6}, A2 = Na = {a, a7 |
}, A3 = Na2 = {a2, as}, A4 = Na3 = {a3, a9}, As = Na4 = {a 4, a!O}, A6 = Nas = {as, all}. Now M = {1, a2, a4, a6, as, a!O}. Hence MIN consists of the elements Av A3 and A 5• MIN <l GIN, and we can talk of (GIN)/(MIN). The elements of this group are the cosets of MIN in GIN. These cosets are Multiplication is calculated in the usual way, B2B2 = (MIN)A 2(MIN)A 2 = (MIN)(A 2A 2) Now the product of A2 and A2 is also the product of cosets. As A2 = Na, A2A2 = Na2 = A 3. Hence BzS2 = (MIN)A3 = B!. It is clear then that (GIN)/(MIN) = {Bv B 2} = gp(B2) is a cyclic group of order 2. Now let us decide what GIM is. The elements of GIM are the cosets C! = M = {1,a2,a4,a6,aS,a!O} and C2 = Ma = {a,a3,a5,a7,a9,all} Clearly GIM = gp(C2) is cyclic of order 2. Hence GIM "'" (GIM)/(MIN). 4.92. Let G be any group. Let F! (G) be all possible nonisomorphic factor groups of G. Let Fi+! (G) = In the particular case that G is cyclic of {All possible factor groups of the groups in Fi (G)}. order 25, find all nonisomorphic groups in U Fi(G), i=l i.e. FdG)uF2 (G)u···. (Hard.) 00 Solution: It is sufficient to consider FtlG). For if L E F 2(G), L = MIN where ME F! (G). But then M = GIK, by definition of F! (G). Thus L is a factor of a factor group. Hence by the factor of a factor theorem it is isomorphic to a factor group of G. We must therefore find the number of factor groups |
of G. All subgroups of G are known. G has unique subgroups of orders 1,2,22,23,24 and 25 by Theorem 4.9, page 105. The factor groups will therefore be of orders 25,24,23,22,2 and 1. In each case the factor groups will be cyclic. For if G = gp(a) and N is a subgroup of G, gp(Na) = GIN. Hence the result. c. The subgroup isomorphism theorem In the homomorphism theorem we were able to say that the image of a homomorphism (): G ~ K was essentially a factor group of G. What can we say about the effect of () on subgroups? Let H be a subgroup of G. Let (h = 8rH, i.e. 8! is the mapping of H to K de fined by h8! = h8, hE H. Then 8! is a homomorphism of H ~ K, and so H8! = H8 e. HI(Ker8!). Now if Ker8 = N={xl xEG, x8 = I}, then Ker8! = {xl xEH and X(}l = x(} = I} = HnN. So H8 = H(}! e. HI(HnN). On the other hand, we know H8 is a subset of G(} and G(} "" GIN. Our question is: what has HI(HnN) got to do with GIN? It must be isomorphic to some subgroup of GIN. But which? This is what the subgroup isomorphism theorem is about. Sec. 4.4] HOMOMORPHISM THEOREMS 125 Theorem 4.23 (Subgroup Isomorphism Theorem, also called the Second Isomorphism Theorem); Let N <J G and let H be a subgroup of G. Then HnN <J H, HN is a subgroup of G, and (HN = {hn I h E H, n E N}) H/(HnN) "" HN/N Proof: If n E HnN and hE H, then h-Inh EN as N <J G, and h-Inh E H as n E H. Therefore h-Inh E HnN and HnN <J H. |
HN is a subgroup, for it is not empty; and if Xl, X2 E HN, then Xl = hlnl, X2 = h2n2 and XIX;1 = hlnln;lh2- 1 = hln3h;1 = hlh;lh2n3h;1 = hn4 where n3 = nln;l EN, hlh;l= hE Hand h 2n 3h;l= n4 EN as N <J G. Hencexlx;l E HN and HN is a subgroup of G. Let cp: H ~ HN/N be defined by hcp = Nh. Then cp is clearly onto, i.e. Hcp = HN/N. Also cp is a homomorphism: (hlh'J)cp = N(hlh2) = NhlNh2 = hlcph2cp. By the homomorphism theo rem, Hcp "" H/(Kercp). Kercp = {x I X E H, Xcp = 1} = {x I x E H, Nx = N}. If Nx = N, then 1x = x EN; and if x EN, Nx = N. Therefore Kercp = {x I x E H, x EN} = HnN. Hence HN/N "" H/(HnN). Problems 4.93. Let Q* be the multiplicative group of rationals. Let N = {1, -1}. Let H be the subgroup generated by H-}. Find HN, HN/N and thereby verify the assertion of the subgroup isomorphism theorem that HN/N"" H/HnN. Solution: The elements of H are all of the form (t)r, r various integers. HN = {x I x = hn, hE H, n E N} = {x [ x = h or x = -h, h E H} = {x [ x = ±(i)r for all integers r} A coset of HN/N is of the form Nx = {1, -1}x = {x, -x} where x E HN. Now if x E HN, x = ±(i)r. Hence each coset is of the form {(i)r |
, -(i)r}. Since N(i)· N(i) •...• N(!) = N(i)r and '----==------=--" r'(!)r E N(i)r, each coset of HN/N is a power of N(i). Thus gp({N(!)}) = HN/N; and since (!)r e N for r =F 0, HN/N is the infinite cyclic group. Now HnN = {x I x = (-~)r for some r and x = ±1} = {1}.Hence H/(HnN) "" H. But H is infinite cyclic. Thus we have verified that H/(HnN) "" HN/N. 4.94. Let u = ( 1 2 3 2 1 3 ::: : ) and H = gp({u}). Prove that HAn/An is cyclic of order 2. Solution: HAn/An"" H/(HnA n). Now u is an odd permutation, hence u e An. Also H = {u,,}, so HnA n = {,}. Therefore HAn/An"" H/{,} "" H. But H is cyclic of order 2. Thus HAn/An is cyclic of order 2. 4.95. Let a group G contain two normal subgroups M and N. Let H be a subgroup of G. Prove that HM/M "" HN/N if HnM = HnN. Solution: By the subgroup isomorphism theorem, HM/M H/HnM H/HnN "" HN/N. Hence HN/N "" HM/M by Problem 4.8, page 97. 4.96. If in the preceding problem we know that G/M has every element of order a power of 2, show that H/(HnN) has every element a power of 2. Solution: H/(HnN) = H/(HnM) "" HM/M <: G/M. Hence the result. 126 ISOMORPHISM THEOREMS [CHAP. 4 4.97. Let Hand K be subgroups of G, N <J G and HN = KN. Prove that HI(HnN) c= KI(KnN). Solution: HI(HnN) == H |
NIN = KNIN == KI(KnN). Then by Problem 4.8, page 97, HI(HnN) cc KI(KnN). 4.98. Let G d G I d {I} and G I <J G. Suppose GIGI and G I are abelian and H is any subgroup of G. Prove that there exists a subgroup HI of H such that HI <J H, and HIHI and HI are abelian. (Hard.) Solution: Let HI = HnG I. Then by the subgroup isomorphism theorem, HI <J H, and HIHI = HGI/G I. But HGI/G I C GIG I and GIGI is abelian. Therefore HIHI is abelian. As HI C G I and G I is abelian, we conclude that HI is abelian. 4.99. Let G d G I d G2 d {I}. Let G I <J G, G2 <J G I, and suppose GIGI, GI /G 2 and G2 are abelian. Prove that if H is any subgroup, then it has subgroups HI and H2 such that HI <J H, H2 <J HI and HIH I, HIIH2 and H2 are abelian. (Hard.) Solution: Let HI = HnG I. Then, as in Problem 4.95, HI <J Hand HIHI is abelian. Now consider HI as a subgroup of G I • As G2 <J G I, by the subgroup isomorphism theorem H l nG 2 <J HI and H I /(H l nG 2 ) == HIG2/G2CGI/G2' Since GI /G2 is abelian, so is H I /(H l nG2). Con sequently we put H 2 = HI n G2. Finally as H2 C Gz and G2 is abelian, so is Hz and the result follows. d. Homomorphisms of cyclic groups We return now to study cyclic groups. In Section 4.2b we could state that in a finite cyclic group there was at most one subgroup of any given order. An analogy for the infinite cyclic group would have been awkward to formulate without the concept of index which we have had at our disposal since Section 4 |
.3a. Recall that a subgroup H of a group G is of index n in G if there are exactly n distinct In the case of finite cyclic groups we have proved that there is one right co sets of H in G. and only one subgroup H of any order m dividing [G[. Since [G: H] = [G[lm, there is one and only one subgroup of any given index dividing the order of G. This gives the clue to Theorem 4.24: There is one and only one subgroup of any given finite index n> 0 in the infinite cyclic group. Proof: Let G = gp(x) where G is infinite cyclic. Let Hn = gp(xn). Then Hn = {xnr [ all integers r}. Hence the co sets H n, H nX,..., H nXn- 1 are all distinct and are all the cosets of Hn in G. Hence Hn is of index n. Next if H is a subgroup of index n, we already know that H is generated by the smallest positive power xr E H (Theorem 4.9, page 105), and this means H = Hr. But Hr is of index r. Hence r = nand Hr = Hn. This concludes the proof. With our knowledge of cyclic groups it is easy to apply the homomorphism theorem to find all homomorphisms of cyclic groups. Theorem 4.25: Let B be a homomorphism of a cyclic group G. Then GB is cyclic; and if [G[ < 00, [GB[ divides [G[. Furthermore if H is any cyclic group such that [H[ divides [G[, there is a homomorphism of G onto H. If G is infinite cyclic, there is a homomorphism of G onto any cyclic group. Proof: If G = gp(x), then G = {xr [ all integers r}, and Go = {xrB [ all r} = {(xB)r [ all r} = gp(xO) CHAP. 4] SUPPLEMENTARY PROBLEMS 127 Thus GB is cyclic. If IGI < 00, then by the homomorphism theorem GB ~ GIN for some normal subgroup N of G. Hence IGBI = IGI/INI (see Lagrange's theorem) and IGBI divides |
IGI. Suppose H is cyclic, H = gp(y), and the order m of H divides the order of G = gp(x). Say, IGI = rm. Let N = gp(xm). Then N = {I, xm,..., (Xm),-l}, INI = r, and N is of index m. Because G is abelian, GIN makes sense and is of order m. But G is cyclic and conse quently so is GIN. Hence GIN ~ H, and H is a homomorphic image of G. If G is infinite cyclic, then, as we saw in the proof of Theorem 4.24, [G: Gnl = n if Gn = gp(xn). As Gn <l G, GIGn is of order n. But GIGn is the homomorphic image of a cyclic group; hence it is cyclic. Thus G has as homomorphic image any cyclic group of order n, n > O. Ob viously it also has the infinite cyclic group as a homomorphic image. A look back at Chapter 4 In this chapter we have thoroughly investigated the simplest class of groups, the cyclic groups. We know that there are cyclic groups of all orders, we know their subgroups, we know that they have as homomorphic images only cyclic groups, and we know whether any cyclic group G has as homomorphic image a given cyclic group. Furthermore the sub groups of cyclic groups are again cyclic. We have also introduced the concept of coset. The cosets form a partition of the group. Using this fact we obtained Lagrange's theorem which states that the order of a subgroup divides the order of a finite group. This enables us to eliminate certain groups as possible subgroups of a given group. We will see later on that it also enables us to find more quickly the groups of a given order. Next we have introduced the idea of a normal subgroup. This gives rise to a new way of looking at homomorphisms, namely as factor groups (see the homomorphism theorem). The subgroup isomorphism theorem tells us that the subgroup corresponding to a given subgroup H of G in a factor group GIN is isomorphic to a factor group of H itself, namely HI(HnN). The factor of a factor theorem tells us that a factor group of a factor group GIN is just |
a factor group of G of the form GIM. Finally, the correspondence theorem associates with each subgroup of the image of a homomorphism B: G ~ K a unique subgroup of G itself. Supplementary Problems FUNDAMENTALS 4.100. Prove that if a and b are elements of a group G and if a- 1b2a = ba, then b = a. 4.101. Suppose a and b are elements of a group G. If a2 = 1 and a- 1b2a = b3, prove b5 = 1. (Hard.) 4.102. Suppose a and b are elements of a group G. If a- 1b2a = b3 and b- 1a2b = a3, prove a = 1 = b. (Very hard.) 4.103. Suppose G and H' are groups. Suppose that G cannot be generated by two elements but that H can. Prove G and H are not isomorphic. 4.104. Let X be a non-empty set and let Y = {y} be disjoint from X. Prove Sx ~ S xu y if and only if X is infinite. 128 ISOMORPHISM THEOREMS [CHAP. 4 CYCLIC GROUPS 4.105. Let G be a cyclic group. Prove that if N is a subgroup of G such that GIN == G, then N = {I}. 4.106. Let G = Z X Z where Z is the set of integers. Define a binary operation 0 in G by (k, l) 0 (m, n) = (k + m, l + n) where (k, l), (m, n) E G. G is a group with respect to this composition. Prove that G is not cyclic. 4.107. Let G1, G2, ••• be subgroups of a group G. If Gi C; Gi + 1> Gi # Gi + 1 for i = 1,2,..., prove that G 1 U G2 U ••• is not a cyclic group. (Hard.) 4.108. Let G be a group and let e: G..... F be a homomorphism. Let C be a cyclic subgroup of G. Let ee E C for all e E C. If H is any subgroup of C, prove that he E H for all h E |
H. 4.109. Let Q be the additive group of rationals with respect to addition. Prove that every two-generator subgroup (# 0) of Q is infinite cyclic. COSETS 4.110. Let H be a subgroup of G. Prove that e: Hg..... g-lH is a matching of the right co sets of H in G with the left cosets of H in G. 4.111. Let Hand K be subgroups of a group G. Show that a coset of H intersection a coset of K is a coset of HnK. 4.112. Let D be the group of Problem 3.72, page 91. Let N = {(O,z) I z E Z}. Prove that N <J D and DIN is infinite cyclic. 4.113. Let G be the group of Problem 3.74, page 91. Let N = {(O, q) I q E Q}. Prove that N <J G and GIN is infinite cyclic. Show that N is isomorphic with the additive group of rationals. 4.114. Let W be the group of Problem 3.77, page 91. Let M = {(O, b) I (0, b) E W}. Show that M <J G and that GIM is infinite cyclic. 4.115. Let G be a group, let H be a subgroup of G, and let g be an element of G. Prove that if N(H) is the normalizer of Hand N(g-lHg) the normalizer of g-lHg, then g-lN(H)g = N(g-lHg). g-lHg = {g-lhg I hE H}. HOMOMORPHISM THEOREMS 4.116. Let q = { ( :!) I a, b, e, d E z}. Prove that q forms a group with respect to the operation + defined by Let e: q ~.Z be defined by (: ~) e = a + d. Prove that e is a homomorphism of q onto the additive_group of integers and find its kernel. Consider q/(Ker e) and prove that in accordance with the homomorphism theorem it is isomorphic with the additive group of integers. 4.117. Let q = {(: |
:) I ad - be oF 0, a, b, e, d real numbers} with operation matrix multiplication. Let e : q..... R*, the nonzero real numbers, be defined by (: :) e = ad - be. Prove that e is a homo morphism from q onto the mUltiplicative group of nonzero real numbers and find its kernel. Prove that q/(Ker e) == R* in accordance with the homomorphism theorem. CHAP. 4] SUPPLEMENTARY PROBLEMS 129 4.118. Let G be any subgroup of Sn, the symmetric group of degree n. Let 8: G..... {1, -1} be the mapping defined by X8 = 1 if x is an even permutation and x8 = -1 if x is an odd permutation. Prove that 8 is a homomorphism of G into the group {1, -1} with operation multiplication of integers. Using the homomorphism theorem, prove that the even permutations of G form a normal subgroup of G. 4.119. Let G be a group and N a normal subgroup of G. Suppose that H = GIN has a sequence of sub i=1,2,...,n-1. Prove that '" d Gn such that [Gi : Gi+ 1] = i + 1, i = 1,..., n-1. groups H = H 1d H 2d... dHn where [Hi: H i+ 1] =i+1, for G has a sequence of subgroups G = G1 d G2 d 4.120. Let M and N be normal subgroups of G with M d N. Prove that GIN is finite if GIM and MIN are finite. 4.121. Let G be a group and N a normal subgroup of G. Suppose GIN has a factor group which is infinite cyclic. Prove that G has a normal subgroup of index n for each positive integer n. 4.122. Let G be a finite group with normal subgroups M and N. Let H be a subgroup of G. Suppose that the orders of M and H and those of Nand H are co-prime. Prove that HMIM == HNIN. 4.123. Let N,M be normal subgroups of G, NdM. Suppose GIN is cyclic and \NIM\ = |
2. Prove that GIM is abelian. 4.124. Find a group G with normal subgroups Nand M, N d M, GIN cyclic, NIM cyclic but GIM not abelian. Chapter 5 Finite Groups Preview of Chapter 5 The most important result of this chapter is a theorem of Sylow which guarantees the existence of subgroups of prime power order. We prove two other theorems of Sylow con cerning subgroups of prime power order and then examine groups of prime power order. One result is that groups of prime power order always have non-trivial centers. In order to construct a new group from any two groups G and H, we define a binary operation on the cartesian product of G and H. The resultant group is called the direct product of G and H. A simple condition enables us to conclude that a group is a direct product. The concept of direct product together with general theorems about subgroups, e.g. the Sylow theorems, help us to classify finite groups. In this chapter we find all groups up to order 15. We study a class of groups called solvable groups. Solvable groups are used in Galois theory to determine whether an equation is solvable in terms of nth roots. An ambitious plan for studying finite groups is to find all simple groups, i.e. groups without proper normal subgroups, and then see how groups are built from simple groups. The Jordan-Holder theorem shows that in a sense a group is built from simple groups in only one way. As yet the task of finding all simple groups is far from complete. We con clude the chapter by exhibiting a class of simple groups, namely An, for n ~ 5. 5.1 THE SYLOW THEOREMS a. Statements of the Sylow Theorems Lagrange's theorem (Theorem 4.11, page 109) tells us that the order of a subgroup divides the order of a finite group. Conversely one might ask: if G is a finite group and n I IGI, is there always a subgroup of order n? The answer to this question is no: A4 is of order 12 but has no subgroup of order 6 (see Problem 5.1 below). The following important theorem, however, ensures the existence of subgroups of prime power order. In the following p will denote a prime. Theorem 5.1 (First Sylow Theorem): Let G |
be a finite group, p a prime, and pT the highest power of p dividing the order of G. Then there is a subgroup of G of order pT. Suppose H is a subgroup of G of order a power of a prime p, and IHI is the highest power of p that divides IGI. Then H is called a Sylow p-subgroup of G. By Theorem 5.1 every finite group has a Sylow p-subgroup. In general a group of order a power of the prime p is called a p-group. A Sylow p-sub group H of a group G is a maximal p-group in G, i.e. if He Fe G where F is a p-group, then F = H (see Problem 5.4). lRO Sec. 5.1] THE SYLOW THEOREMS 131 As an illustration we find the Sylow p-subgroups of the symmetric group S3 on {I, 2, 3}, for p = 2 and 3. The elements of S3 given in Section 3.3a, page 57, are (~ ~ ~) U 2 = (~ ~ :) '1'2 = (~ ~ ~) U I = (~ ~ ~) '1'1 = (~ ~ : ) The order of any Sylow 2-subgroup is 2 and the order of any Sylow 3-subgroup is 3, since IS31 = 6 = 2· 3. Now Ti = T~ = T~ = t, so the sets {t, T } are all subgroups of order 2 and therefore, by definition, Sylow 2-subgroups of S3' There are no other Sylow 2-subgroups of S3 because ui = 0'2' O'~ = 0'1 implies S3 has no other elements of order 2. {t, 0'1' O'z} is the only subgroup of order three in S3' so it is the only Sylow 3-subgroup of S3' '1'3 = (~ ~ ~) } and {t, T }, {t, T 3 2 I Theorem 5.2 (Second Sylow Theorem): If H is a subgroup of a finite group G and H is a p-group, then H is contained in a Sylow p-subgroup of G. Two subgroups Sand T of a group G are called conjugate if there is agE |
G such that g-ISg = T. Recall g-iSg = {g-I sg I s E S}. Theorem 5.3 (Third Sylow Theorem): Any two Sylow p-subgroups of a finite group G are conjugate. The number Sp of distinct Sylow p-subgroups of G is congruent to 1 modulo p and Sp divides IGI. (sp is congruent to 1 modulo p if Sp = 1 + kp for some integer k.) Before proving the Sylow theorems, we will use them to show that, up to isomorphism, there is one and only one group of order 15. If IGI = 15 then, by Theorem 5.1, G has at least one subgroup of order 3 and at least one of order 5. Now Theorem 5.3 implies that there are S3 = 1 + 3k subgroups of order 3 and s31 IGI. But (1 + 3k) I 15 implies k = O. Therefore G has one and only one subgroup of order 3. Similarly G has one and only one subgroup of order 5. These subgroups must be cyclic (Problem 4.48, page 110). Let HI = {I, a, a 2 the subgroup of order 5. H l nH2 = {I}, because an element #1 cannot have order 3 and 5 simultaneously. We look at the order of ab in G which must be either 1, 3, 5 or 15. If the order of ab is 1, then ab = 1 and a = b -I which is impossible, for HI n H 2 = {I}. If the order of ab is 3, then gp(ab) = HI, since HI is unique. In this case ab = ai (i = 0, 1 or 2) and b = a i - I which is impossible. If the order of ab is 5, gp(ab) = H 2 • Hence ab = bi (i = 0, 1,2,3 or 4) and a = bi - I which is impossible. Therefore the order of ab is 15 and G is the cyclic group of order 15 generated by abo } be the subgroup of order 3 and H2 = {I, b, b2, b4, b3 } Further applications of the Sylow theorems are given in the problems below and in Section 5.3 |
. Problems 5.1. Show that the alternating group A4 has no subgroup of order 6. Solution: The elements of A4 were given in Section 3.3c, page 62. We repeat them here for convenience:, = 7'1 7'2 2 3 G 2 3 :) G 2 3 :) G 2 3 :) 3 4 4 2 7'4 = 7'3 7'5 2 4 G 2 3 ~) G 2 3 :) G 2 3 ~) 2 1 4 3 7'6 7'7 = 7's = 1 3 C 2 3!) G 2 3 ~) G 2 3 ~) 3 1 1 2 "2 "5 "s 4 1 2 3 :) = (~ = (~ ~) =G 2 3 :) 2 3 3 2 1 4 132 FINITE GROUPS [CHAP. 5 Suppose that A4 has a subgroup H of order 6. o~ = o~ = O"~ = t, so 02' 05 and 0g are of order 2. j = 1,2,...,8. Hence the elements of order 2 are 02,05 and 0g, and the Also Tj is of order 3 for elements of order 3 are TI, T2'..., Tg. Now H is of order 6, so it contains a subgroup of order 3, t, TV Ti = T2 E H. H must also by Theorem 5.1. Therefore Ti E H contain an element of order 2, by Theorem 5.1. Hence H contains a 0i' say H contains 02' Because if H contains 02 it also contains T4' T! = T3, Tg and T~ = T7' This would 02Tl = T4 and TI02 = Tg, IHI = 6. Thus mean H has at least 8 distinct elements, which contradicts the assumption that 0"2 \l H. A similar argument shows 05 and 08 \l H. This means that H does not contain subgroups of order 2, contradicting Theorem 5.1. Therefore our initial assumption is invalid and A4 does not contain a subgroup of order 6. for some i, say TI E H; then 5.2. Find all Sylow p-subgroups of A4 for p = 2 and 3. Solution: The elements of A4 are given in Problem 5.1. The order of a Sylow 2-subgroup is 4, since 22 is the highest power |
of two dividing 12, the order of A 4• Consequently by Lagrange's theorem none of the T'S can be elements of a Sylow 2-subgroup because they are all of order 3 (see Problem 5.1) and 3 does not divide 4. Now 0iOj = Ok where i, j, k E {2, 5, 8} and 0T = t for i = 2,5,8. Hence P = {t, 02' 05' og} is a subgroup of A4 of order 4. P is the only possible Sylow 2-subgroup as there t, 0"2, 0"5' 08' and these elements are in P. The are only four elements having order dividing 4, viz. order of a Sylow 3-subgroup is 3. The sets {t, TI, Ti}, {t, T3, 7"~}, {t, 7"5, 7"~} and {t, 7"7' 7"~} are all sub groups of order 3. These are all the possible Sylow 3-subgroups, as they include all the elements of order 3. Alternately we may use Theorem 5.3: 83 = 1 + 3k must divide 12. Clearly k #- 0 (we already have four subgroups); and if k > 1, 83 does not divide 12. Hence k = 1 and there are exactly four Sylow 3-subgroups. 5.3. If H is a subset of a group G and g E G, then Ig-IHgl = IHI, where g-lHg = {g-lhg I hE H}. Solution: We define a matching a: H ~ g-IHg by a: h -+ g-lhg for hE H. a is clearly an ooto mapping. To show a is also one-to-one, we must prove hi = h2 (hI' h2 E H) if and only if g-lhlg = g-lh2g. Let hi = h 2• Then by multiplying on the left by g-l and on the right by g we get implies hI = h 2. Hence a is a matching and g-lhlg = g-lhzU. Similarly g-lhlg = g-lh2g 1U-1Hgl = IHI· 5.4. Let p |
-group such that P ~ H ~ G, then H = P. IGI = pTm (r""o 1 and p}' m) and let P be a Sylow p-subgroup of G. Prove that if H is a, Solution: Suppose IPI = pT. Hence and t = rand IPI = IHI, and so P = H. IHI = pt, t "" O. By Lagrange's theorem, pt I pTm. Since p f m, t "'" r. But P ~ H 5.5. If H is a Sylow p-subgroup of G, then g-IHg is also a Sylow p-subgroup of G. Solution: Suppose IGI = pTm (r ""0 0 and p J m); then IHI = pT. But Ig-IHgl = IHI by Proble~ 5.3. Hence g-IHg is a Sylow p-subgroup of G if it is a subgroup. To prove g-IHg is a subgroup, observe that (g-lh lg)(g-lh2g)-1 = g-lhlh;-lg E g-IHg. From Lemma 3.1, page 55, g-IHg is therefore a subgroup. 5.6. Prove that a finite group G is a p-group if and only if every element of G has order a power of p. Solution: IGI = pr If then, as every element of G must have order dividing the order of the group, every element has order a power of p. To prove the converse let every element of G have order a power of p and assume the order of G is not a power of p. Then there is some prime q, q #- p, such that q I IGI. But by the first Sylow theorem, G has a subgroup H of order a nonzero power of q. So H contains an element g #- 1. By Lagrange's theorem, the order of g is a nonzero power of q and hence the order of g is not a power of p. This contradicts the assumption that all elements have order a power of p. Hence IGI = pT for some r ""0 O. Sec. 5.1] THE SYLOW THEOREMS 133 5.7. If G has only one Sylow |
p-subgroup H, then H <l G. Solution: If g E G, g-lHg is a Sylow p-subgroup by Problem 5.5. But G has only one Sylow p-subgroup. Thus g-lHg = H for all g E G and H <l G. 5.8. 5.9. If IGI = pq, where p and q primes and p < q, then G has one and only one subgroup of order q. Furthermore if q # 1 + kp for any integer k, then G is the cyclic group of order pq. Solution: By Theorem 5.3 G has 8 q = 1 + kq Sylow q-subgroups of order q, with k"" O. Also 1 + kq divides pq. There can only be four possibilities for 1 + kq as the expression of pq as a product of primes is unique: 1 + kq = q or 1 + kq = p or 1 + kq = pq or 1 + kq = 1. As q does not divide 1 + kq, we are left with the possibilities that 1 + kq = p or 1 + kq = 1. Since q > p, 1 + kq # P and hence k = O. Thus there is only one subgroup of order q, say H. There are 8 p = 1 + kp subgroups of order p. Again we have the possibilities 1 + kp = 1, 1 + kp = p, 1 + kp = q, or 1 + kp = pq as 8 p divides IGI. Clearly p does not divide 1 + kp, so 1 + kp = 1 or 1 + kp = q. The last is not true by assumption, so again there is only one subgroup K of order p. It follows from Problem 5.7 that H <l G, K <l G. Also HnK = {I} as the nonunit elements of H are of order q and those of K are of order p. If hE Hand k E K (h # 1, k # 1), then h -l(k-Ihk) E H as H <l G (h-1k-1h)k E K as K <l G Hence h-Ik-1hk = 1 and hand k commute. By Lagrange's theorem |
, the order of hk is p, q or pq. But (hk)p = hpkp as hand k commute, so (hk)q = k q # 1. Therefore hk is of order pq, and so G is cyclic. (hk)p = hp # 1. Similarly Instead of the argument of the last paragraph, we note that H = {I, h, h 2, •••, hq-I}, K = - I}. Now hk is an element of order 1,p,q or pq. If hk is of order p, since there is {1,k,k 2, •••,kp only one subgroup of order p, gp(hk) = K, i.e. hk = ki for some i, 0 "" i "" p - 1. But then hE K, which contradicts H nK = {I}. Similarly gp(hk) is not of order q, nor of order 1. Thus gp(hk) is of order pq, and so G is cyclic. Show that if q = 1 + kp in Problem 5.8, G is not necessarily cyclic. Solution: Consider S3' the symmetric group on {I, 2, 3}. cyclic group. IS31 = 6 = 3·2, 3 1 + 1 • 2 and S3 is not a 5.10. If IGI = 2p, p an odd prime, then G has one and only one subgroup of order p and either G has exactly p subgroups of order 2 or it has exactly one subgroup of order 2. Solution: From Problem 5.8 we know G has one and only one Sylow p-subgroup. Because p is itself the highest power of p dividing IGI, the Sylow p-subgroup of G is of order p. Thus there is precisely one subgroup of G of order p. The number of Sylow 2-subgroups of G is 82 = 1 + k2 for some integer k. Again 1 + 2k = 1,2, p or 2p. As 2 does not divide 1 + 2k, then 1 + 2k = 1 or 1 + 2k = P and the number of Sylow 2-subgroups is either 1 or p. b. Two lemmas used in the proof of the Sylow theorems Lemmas 5.4 and 5. |
6 will provide the tools for proving the Sylow theorems (see Section 5.1c). Throughout this section G will be a fixed group and H a subgroup of G. As usual we denote subsets of G by A, E, C, etc. A generalization of the concept of normalizer as defined in Section 4.3d, page 112, will be essential. We point out once more that if A is a non-empty subset of a group G and g E G, then g-IAg = {g-Iag I a E A}. Definition: Let A be a non-empty subset of a group G. The set {h I h-IAh = A, hE H} is called the normalizer of A in H and is written N H{A). 134 FINITE GROUPS [CHAP. 5 It is easy to prove that NH(A) is a subgroup of H (Problem 5.11). When H = G, Nc(A) is the normalizer of A as defined in Chapter 4. Definition: Let A and B be non-empty subsets of G. B is said to be an H-conjugate of A if h-iAh = B for some hE H. (Note that if H = G, then A and B are con jugate as defined in Section 5.la.) The next lemma gives us a formula for calculating the number of distinct subsets of G which are H-conjugates of A. Lemma 5.4: If G is a finite group with subgroup H and non-empty subset A, the number of distinct H-conjugates of A is the index of NH(A) in H, i.e. [H: NH(A)]. P1'00f: Since [H: NH(A)] is the number of distinct right co sets of NH(A) in H, we need only define a one-to-one mapping, (t, of the right co sets of NH(A) in H onto the distinct H-conjugates of A. Let (t be defined by ( t : NH(A)h -+ h-iAh (h E H) To show that (t is a one-to-one mapping, we must prove that for hi, h2 E H, NH(A)hi = NH(A)h2 if and only if h;lAhi |
= h;:lAh2 (i) (ii) Let h;lAhi = h;:lAh2. Then A = hih;:lAh2h;1 = (h2h;1)-lA(h2h;1). Hence h2h;1 E NIJ(A) and so h2 E NH(A)hi. Since two right cosets are equal or disjoint, we conclude NH(A)h i = N H(A)h2. Thus h;lAhi = h;:lAh2 implies NH(A)hi = N H(A)h2. then hi E NfJ(A)h2, If NH(A)hi Therefore for some n E NH(A). i.e. hi = nh2 NH(A)h2, because n-iAn = A by definition of NH(A). Hence NH(A)hi = NH(A)h2 (t is clearly onto, so the proof is complete. h;lAhi = h;:lAh2. Most of our arguments are concerned with sets whose elements are subsets of G. We implies denote such sets by script letters eA, CJ3, etc. For example, let G be the cyclic group of order 6, G = {I, a,..., a5 }. Subsets of G are, for example, A = {l,a}, B= {a2,a3,a6 }, C= {a}. An example of a set whose elements are subsets of G is the set whose elements are A and B. We write eA = {A, B}. Another such set would be CJ3 = {A, B, C}. Proposition 5.5: Let eA be a set of subsets of G. We define for A,B E eA, A - B if B is an H-conjugate of A (i.e. if there exists an element hE H such that is an equivalence relation on eA (see Problem 5.16 h -lAh = B). Then for the proof). We will make a few observations about -, which follow because it is an equivalence relation on eA. Recall that if A E eA, A - = {X I X E A and X - A}, i.e. A is the equivalence class containing A ( |
see Section 1.2c, page 9). Recall that the distinct equiva lence classes are disjoint and that their union is eA (Theorem 1.2, page 10). By a set of representatives of the equivalence classes we mean a set '1( which contains one and only one element from each of the distinct equivalence classes. It follows that eA is the disjoint union of the sets R -, R E '1(. Hence leAl = ~ IR -I. We are now in a position to prove our main lemma. R E ~ Lemma 5.6: Let eA (~~) be a set of subsets of G. Suppose that for each A E eA and each hE H, h-iAh E eA. Let - denote the equivalence relation defined by A - B if B is an H-conjugate of A. Let '1( be a set of representatives of the equivalence classes. Then leAl Sec. 5.1] THE SYLOW THEOREMS 135 Proof: We know from the remarks above that leAl ~ IR-I RE<,R. But R - = {X I X = h-1Rh for some hE H} since h-1Rh E eA for every hE H. So is the set of H-conjugates of R. The number of such H-conjugates is, by Lemma 5.4, R [H: NH(R)]. Hence leAl = ~ [H: NH(R)], as claimed. R E <,R. Corollary 5.7: Let P (# 0) be a subset of G. Let eA = {g-lPg I g E G}. Let CJ{, Hand - be as in Lemma 5.6. Then leAl = ~ [H: NH(R)] = RE<,R. [G: Nc(P)] Proof: Clearly [eAl is the number of G-conjugates of P, and the result follows from Lemma 5.4. Corollary 5.8: Let eA = {A I A is a subset of G and A has precisely one element}. Let -- be the equivalence relation in eA when H = G, and let CJ{ be a set of representatives of the equivalence classes. Let CJ{* = (R I RnZ(G) = 0, R |
E CJ{}. Then IGI = IZ(G)I + ~ [G: Nc(R)] R E <,R.* (We remind the reader that Z(G) = (x I xg = gx for all g E G}) Proof: Clearly leAl = IGI; hence IGI ~ [G: Nc(R)] RE<,R. (5.1) If Z E Z(G), then {z} E eA and the number of G-conjugates of {z} is one, namely {z} itself. Consequently {z} E CJ{ for each z E Z(G). Note that Nc({z}) = G if z E Z(G). Hence adding first the contribution made by all R E CJ{ with R nZ(G) # 0 in (5.1), we obtain [Z(G)[ and the result follows. Note that as R = {r}, Nc(R) {g I g E G and g-lrg E R} {g I g E G and g-lrg = r} C(R) (For the definition of C(R), the centralizer of R in G, see Section 4.3d, page 112.) Hence Corollary 5.8 takes the form (5.2) is called the class equation of G. IGI = IZ(G)I + ~ [G: C(R)] R E <,R.* (5.2) Problems 5.11. If A is a non-empty subset and H a subgroup of a group G, then NH(A) is a subgroup of G. Solution: NH(A) is clearly a subset of G. NH(A) =1= 0, since 1 E Hand 1- 1A1 = A implies 1 E NH(A). implies nAn- 1 = A, or (n-1)-lAn- 1 = A. Furthermore n- 1 E H, Let n E NH(A). n-1An = A since H is a subgroup; hence n- 1 E NH(A). If m,n E NH(A), (mn)-lA(mn) = n-1(m-1Am)n = n-1An = A; hence mn E NH(A). Accordingly NH(A) is a subgroup of G. 136 5.12. 5. |
13. FINITE GROUPS [CHAP. 5 " Check Lemma 5.4 by direct computation when G = Sa and (using the notation of Section 3.3a, page 57) A = {TI}' H = {t, T2}' Solution: The H-conjugates of A are t-I{TI}t = {TI}, T2 1{TI}r2 = {ra}. Thus the number of H-conjugates of A is 2. Lemma 5.4 requires that 2 = [H: NH(A)]. But NH(A) = {x I x E H and x-lAx = A} = {t}. Hence [H: NH(A)] = 2, as required. Check Corollary 5.7 when G = DB, the dihedral group of degree 8 (G = {b i, abi, 0 "'" i < 8}, where a2 = bB = 1 and a-Iba = b- I; see page 75, with a = T, b = <72), given P = {a} and H = {I, b2, b4, b6}. Solution: cA of Corollary 5.7 is given by cA = {g-IPg I g E G} = {P I,P2,Pa,P4 } where PI = {a}, lcAl = 4. Using the equivalence relation - of the thus P 2 = {ab 2}, P a = {ab 4 }, P 4 = {ab6}; corollary, P2 - = {h- IP 2h I hE H} = {P2,P4 } For "1( choose one representative from each of the equivalence classes, e.g. choose Then Lemma 5.6 claims that N H(P2)' Hence We must also show that [G: NdP)] = 4. As "1( = {Pa,P2}. lcAl = 4 = [H: NH(Pa)] + [H: N H(P2)]. Now Nu(Pa) = {1,b4 } = [H: NH(Pa)] = [H: N H(P 2 )] = 2 and the required equation of Corollary 5.7 holds. {x I x E G and x-Ipx = P} {x I x E G and x-lax = |
a} {I, a, b4, ab4 } the index of NdP) in G is 4, the required number. 5.14. Check Corollary 5.8 when G = Sa. Use the notation of page 57. Solution: cA (of Corollary 5.8) = {Pl>P2,Pa,P4,P5,P6} where PI = {t}, P 2 = {<71}' P a = {<72}' P 4 = {TI}, P 5 = {T2}' P 6 = {Ta}· Let - be as in Corollary 5.8. Consequently P I - = {PI}' P 2 - = {P 2,Pa}, P 4 - = {P4, P 5, P 6 }. To define "1(, we choose one element from each of these equivalence classes. j = 1. Therefore Let us take "1(* = {P Z,P4}· Now N C (P2) = {t,<71><72} and NdP4) = {t,TI}' Hence, as required, "1( = {PI, P 2, P 4 }· As Z(G) = {t}, PjnZ(G) = 0 except for 5.15. Show that NH(A) = Nc(A)nH for any non-empty subset A and subgroup H of a group G. Solution: Let n E NH(A); then n E Hand n-IAn = A. But H ~ G, so that nEG and by definition nENc(A). Consequently NH(A)~NdA)nH. If nENc(A)nH, then n-1An=A and nEH. Thus Nc(A)nH ~ NH(A) and the equality follows. 5.16. Prove Proposition 5.5, page 134. Solution: As H is a subgroup of G, H contains the identity, so A = 1- I Al and thus A-A. If A - B, such that h-IAh = B. Consequently (h-I)-IB(h- l ) = A and then there is an element hE H such that h-1Ah = Band B - A. Finally, if A - Band B - C, g-IBg = C. |
It follows that hg E Hand (hg)-IA(hg) = g-l(h-1Ah)g = g-IBg = C, and so A-C. Hence - is an equivalence relation on cA. then there exist h,g E H 5.17. Let A, B be subsets of G. Suppose B is an H-conjugate of A, where H is a subgroup of G. Prove that [H: NH(A)] = [H: NH(B)]. Solution: Let cA = {X I X = g-IAg or X = g-IBg for some g E G}. We use Proposition 5.5. IA -I is therefore the number of H-conjugates of A, A - = B -, as B is an H-conjugate of A. and also the number of H-conjugates of B. By Lemma 5.4, [H: NH(A)] = [H: NH(B)]. Sec.5.1J THE SYLOW THEOREMS 137 c. Proofs of the Sylow theorems First we prove a weak form of the first Sylow theorem. Proposition 5.9: If G is a finite abelian group and p is a prime dividing the order of G, then G has an element of order p. Proof: We will prove the proposition by induction on the order of G. If IGI = 1, there is nothing to prove. Assume the proposition is true for all groups of order less than n, the order of G, where n> 1. Recall from Section 4.2b, page 105, that if G is cyclic there is a subgroup of order any integer that divides IGI. Thus if G is cyclic the theorem holds, and we may therefore assume G is not cyclic. If n is a prime, G is cyclic; hence n is not a prime. Suppose h (~1) E G, h of order m. Clearly m < n. Let H be the cyclic group gen erated by h. H is a proper subgroup of G. Now if pi m, by the induction assumption, H has an element of order p. If pIm, form the factor group G/H (every subgroup of an IGIHI = abelian group is a normal subgroup so H <J G). Since IGI/IHI |
, p IIGI/IHI. Therefore by the induction assumption, GIH has an element {j of order p. Let v: G ~ GIH be the natural homomorphism of a group onto its factor group (see Theorem 4.17, page 114) and g be a preimage of {j under v. Now (gp)v = {jP = the identity of GIH, so gP E H. As H is of order m, (gm)p = (gp)m = 1. Therefore gm has order p or gm = 1. If gm = 1, then gmv = {jm = 1. Since {j has order p this implies p divides m, con trary to our assumption. Therefore gm is an element of G of order p. IGIHI < IGI. As IHI > 1, We are now in a position to prove the Sylow theorems. For convenience we repeat the statement of each theorem. The First Sylow Theorem: Let G be a finite group, p a prime and pT the highest power of p dividing the order of G. Then there is a subgroup of G of order pT. Proof: We will prove the theorem by induction on the order n of G. For IGI = 1 the theorem is trivial. Assume n > 1 and that the theorem is true for groups of order < n. Suppose IZ(G)I = c. We have two possibilities: (i) p [c or (ii) p yc. (i) Suppose pic. Z(G) is an abelian group. By Proposition 5.9, Z(G) has an element of order p. Let N be a cyclic subgroup of Z(G) generated by an element of order p. N <J G, IGINI = nip by since any subgroup of Z(G) is normal in G. Consider GIN. Then Corollary 4.14, page 110. Hence by our induction assumption, G/N has a subgroup jj of order pT-l. By Corollary 4.21, page 121, there exists a subgroup H of G such that HIN = H. As pT-l = [ill = [HI/IN[ = IHI/p, we conclude that IHI = pT. Thus in this case, G has a subgroup of order pT |
. (ii) Suppose p)' c. The class equation for G is (see Equation (5.2) of Corollary 5.8, page 135) IG[ = [Z(G)[ + ~ [G: C(R)] RE~. Since p IIGI and pIc, we have Py ~ [G: C(R)]. Therefore for at least one R E ~*, IGI = [G: C(R)]IC(R)I by Corollary 4.14 to Lagrange's theorem, p %[G: C(R)]. But IC(R)I = IGI, page 110. Hence pT IIC(R)[, since pT IIGI. Now then C(R) = G and RnZ(G) = R, contrary to the assumption that RnZ(G) = p. Thus by the induction assumption, C(R) has a subgroup H of order pT. Consequently so does G. IC(R)I ~ IGI; for if R E '1(. In either case we have found a subgroup H of order pT. The proof is complete. 138 FINITE GROUPS [CHAP. 5 The following gives a simple formula for the normalizer of a Sylow p-subgroup P in a subgroup H of G, where IHI is a power of p. It will be used in the proof of the second Sylow theorem. l.. emma 5.10: If G is a finite group, P a Sylow p-subgroup of G, and H is a subgroup of G of order a power of p, then Proof: PnH C NH(P), as conjugation by an element of P sends P to itself. We show NH(P) cPnH. NH(P) CNG(P) and P <J NG(P) (see Problem 5.15 and Problem 4.60, page 113), so that by the subgroup isomorphism theorem (Theorem 4.23, page 125) we have: NH(P)P is a subgroup of G and NH(P)P/P "'" NH(P)/NH(P) nP [N H(P)P : P] = [N H(P) : N H(P) n Pl. But N H(P) is a p-group, i.e. a group |
of Consequently order a power of p, since it is a subgroup of the p-group H. Thus [NH(P): NH(P) np] is a [NH(P)P: P] is therefore also a power of p and, as P is a p-group, INH(P)PI power of p. is a power of p. Accordingly, NH(P)P is a p-group. But Pc NH(P)P and P is a Sylow p-subgroup. Hence P = NH(P)P, for P cannot be a proper subgroup of any other p-sub group of G (see Problem 5.4, page 132). NH(P) is therefore a subgroup of P. As NH(P) C H, we conclude NH(P) C HnP. The Second Sylow Theorem: Let H be a subgroup of a finite group G, and let P be a Sylow p-subgroup of G. If H is a p-group, then H is con tained in a G-conjugate of P. Proof: We apply Corollary 5.7, page 135, to eA = {g-lPg I g E G} to conclude By Lemma 5.10, NH(R) = HnR for each R E CJ?... Hence leAl = ~ [H: NH(R)] = [G: NG(P)] RE~ [G:NG(P)] = ~ [H: HnR] RE~ (5.3) If HnR =1= H for all R E CJ?.., as H is a p-group, the right-hand side of equation (5.3) is divisible by p. Hence [G: NG(P)] is divisible by p. But Pc NG(P), so that p does not divide [G: NG(P)]. This contradiction implies that HnR = H for at least one R E CJ?... But as R E eA, R is a G-conjugate of P. The result follows. The Third Sylow Theorem: (i) Any two Sylow p-subgroups of a finite group G are G-con (ii) The number Sp of distinct Sylow p-subgroups of jugate. G is congruent to 1 modulo p. (iii) Sp IIGI. Proof: |
(i) Let P and pI be two Sylow p-subgroups of G. By the second Sylow theorem, pI, as a p-group, is contained in some G-conjugate R of P. But lP/l = IRI, by Problem 5.3; page 132. Hence pI = R and pI is conjugate to P under G. (ii) Let P be any Sylow p..,subgroup of G. Since any other Sylow p-subgroup is conjugate to P and any conjugate of a Sylow p-subgroup is a Sylow p-subgroup (Problem 5.5, page 132), we conclude by Lemma 5.4 that But on putting P = H in Equation (5.3), we have Sp = [G: NG(P)] Sp = ~ [P: PnR] RE~ Sec. 5.2] THEORY OF p-GROUPS 139 Now for exactly one R E'l{, R = P; for the only P-conjugate of P is P itself and so P is the only possible representative of its equivalence class. In all other cases, P n R # P. Therefore [P: P n R] is a power of p for all R E 'l{ except one, and for this one [P: PnR] = 1. Hence (iii) By Corollary 4.14 to Lagrange's theorem, [G: Nc(P)] INc(P)i. Since Sp [G: Nc(P)], Sp IIGI. sp = 1 + kp IGI 5.2 THEORY OF p-GROUPS a. The importance of p-groups in finite groups Suppose that G is a finite group. In Section 5.1a we saw that G has a Sylow p-subgroup for any prime p. (p will be a prime throughout this section.) One reason why the study of p-groups (groups of order a power of p) is so important is that the structure of the Sylow p-subgroups of G partly determines the structure of G. One instance is the follow ing theorem: If G is a finite group whose Sylow p-subgroups are all cyclic, then G has a normal subgroup N such that GIN and N are both cyclic. (M. Hall, Jr., The Theory of Groups |
, Macmillan, 1959, Theorem 9.4.3, page 146.) In this section we shall determine some of the elementary properties of p-groups. b. The center of a p-group A very important property of finite p-groups is given by Theorem 5.11: If G # {I} and G is a finite p-group, then Z(G), the center of G, is not of order 1. Proof: We make use of the class equation (equation (5.2), page 135) IGI = IZ(G)I + ~ [G: C(R)] R E Cf(* (5.2) It follows immediately from the definition of C(R) and Z(G) that C(R) = G if and only if R C Z(G). Because the sum on the right side of (5.2) is taken over all R such that RnZ(G) = ~ and because IGI = pr, pi [G: C(R)] for all R E'l{*. Hence p I ~ [G: C(R)]. Since p IIGI, we can conclude that p IIZ(G)I, which means Z(G) # {I}. R E Cf(* Corollary 5.12: If G is a group of order pT, order pr-i. r ~ 1, then G has a normal subgroup of Proof: The proof is by induction on r. The statement is clearly true for r = 1. Sup pose the corollary is true for all k < r where r> 1. By Theorem 5.11, Z(G) # {I}. Because p IIZ(G)I, Proposition 5.9 implies Z(G) has an element g of order p. Let N = gp(g). IGINI = pr-i. There N <J G, since any subgroup of Z(G) is normal in G. Consider GIN. fore by the induction assumption, GIN has a normal subgroup fl of order pr-z. By Corol lary 4.21, page 121, there exists a subgroup H of G which contains N and such that IHI = pr-i. Furthermore, again by Corollary 4.21, H <J G. Thus G HIN = fl. Then has a normal subgroup of order pr-i and the proof is complete. Clearly |
we could repeat this argument until we obtain a sequence of subgroups of G {I} = Ho C Hi C... C H r- i C Hr = G (5.4) where Hi <J H i + i (i = 0,1,..., r-1) and IHil = pi (i = 0,1,2,..., r). 140 FINITE GROUPS [CHAP. 5 Problems 5.18. Suppose G is a group with S a subgroup of the center Z(G). Prove G is abelian if GIS is cyclic. Solution: Suppose GIS is cyclic. Then we can find a E G such that every element of GIS is a power of as. Then if g, hE G, g = aiz, h = aiz' for a suitable choice of the integers i and j, with z and z' in S. Then gh = aizaiz' = aiaizz l = aiaiz'z = aiz'aiz kg Thus every pair of elements of G commute. Hence G is abelian. 5.19. Prove that a group of order p2 is abelian (p a prime). Solution: Let G be of order p2 and let Z be the center of G. By Theorem 5.11, Z =F {I}. If Z = G, G is IGIZI = p, so GIZ is cyclic. By Problem 5.18, it follows that G then abelian. Suppose Z =F G; is abelian. 5.20. Let A = {0,1,..., p group. Let I}, where p is a prime. Then under addition modulo p, A is an abelian G = {(a,b,e) I a,b,eEA} be the set of all triples (a, b, c) of elements of A. Define (a, b, c) • (ai, b', e' ) (a + a', b + b', e + e' - ba'l Prove that with respect to this binary operation, G is a non-abelian group of order p3. Solution: It is clear that IGI = p3. To prove that G is a group, we check first the associative law: «a, b, e)(a', b', e' |
)) • (a", b", e") (a + ai, b + b', e + e' - ba')(a", b", e") (a + a' + a", b + b' + b", e + e' + e" - ba' - (b + b')a") On the other hand, (a, b, e)(a', b', e')(a", b", e")) (a, b, e)(a' + a", b' + b", e' + e" - b'a") (a + a' + a", b + b' + b", e + e' + e" - b'a" - b(a' + a")) We check that e + e' + e" - ba' - (b + b')a" e + e' + e" - b'a" - b(a' + a") which is true. Thus G is a semigroup. Now (a, b, c) • (0,0,0) = (a, b, c) (0,0,0) • (a, b, c) and so (0,0,0) is the unit element of G. Finally, (a, b, c) • (-a, -b, -e - ba) = (0,0,0) (-a, -b, -e - ba)(a, b, c) and hence every element of G has an inverse. Our last task is to prove that G is non-abelian. Now (1,0,0)(0,1,0) = (1,1,0), (0,1,0)(1,0,0) (1,1, -1) and thus (1,0,0)(0,1,0) =F (0,1,0)(1,0,0). 5.21. Let A be the additive group of integers modulo p, A = {O, 1,..., p group of integers modulo p2, B = {O, 1,..., p2 j E B. Prove that under the binary operation I}. Let G be the set of all pairs I}; and let B be the additive (i, j), i E A, (i, j) • (ii, j') = (i + ii, j + j' + |
ji'p) G is a non-abelian group of order p3. Solution: Clearly G is of order p3. We check that G is a semigroup. Sec.5.2J THEORY OF p-GROUPS 141 «i, j)(i', j'»(i", j") (i + i', j + j' + ji'p)(i", j") (i +i + i", j + j' + j" + ji'p + (j + j' + ji'p)i"p) (i, j)( (i', j')( i", j"» (i, j)(i' + i", j' + j" + j'i"p) (i + i' + i", j + j' + j" + j'i"p + j(i' + i")p) To prove that the binary operation in G is associative, we need check only that Since p2 = ° in B, this equality is readily verified. ji'p + (j + j' + ji'p)i"p = j'i"p + j(i' + i")p The identity element of G is (0,0). The inverse of (i, j) is (-i, - j + jip). Thus G is a group. Finally, (1,0)(0,1) (1,1), (0,1)(1,0) = (1, 1 + p) and therefore G is non-abelian. 5.22. Prove that the group G in Problem 5.20 has the property that for all g E G, gP = 1 (i.e. (0, a, 0» p is odd. Is this true if p = 2? Solution: if Let (a, b, c) E G. Then (a, b, C)2 = (a, b, c) (a, b, c) (2a, 2b, 2c - ba) Continuing, we find (a, b, c)3 (a, b, C)2 (a, b, c) (2a, 2b, 2c - ba)(a, b, 0) (3a, 3b, 3c - ba - 2ba) By induction it follows that (a, b, c)p = (pa |
, pb, pc - ba - 2ba -... - (p - l)ba) But pa = 0, pb = 0, pc = 0. Finally, ba + 2ba +... + (p - l)ba = p(p 2- 1) ba since 1 + 2 +... + p - 1 = tp(p - 1). If p is odd, p - 1 is even. Therefore!p(p - 1) is an integer l)ba = 0. Thus we have (a, b, c)P = 1. divisible by p. Hence tp(p - If p = 2, then (a, b, C)2 = (2a, 2b, 2c - ba) = (0,0, -ba) In particular if a = 1, b = 1 and c = 0, we have (1,1,0)2 = (0,0, -1). Thus not every element of G is of order 2. This result could have been observed by noting that a group G satisfying g2 = 1 for all g in G is abelian. To see this let g, hE G. Then as (gh)2 = 1, and so G is abelian. But as G is not abelian, not every element is of order 2. gh = (gh)-l = h- 1g- 1 = h 2h- 1g 2g- 1 = hg 5.23. If p is odd, does the group G of Problem 5.21 satisfy gP = 1 (i.e. (0,0» Solution: No, since (0,1)2 = (0,1)(0,1) = (0,2). Inductively, (0, l)P = (0, p) =1= (0,0). for all g E G? 5.24. Prove that if G is a group such that gP = 1 for all g E G, has the same property, i.e. hp = 1 for all h E H. then every homomorphic image H Solution: Let (J be a homomorphism of G onto H. Then if h E H, we can find g E G such that g(J = h. Therefore hp = (g(J)P = (gp)(J = 1(J = 1. 142 FINITE GROUPS [CHAP |
. 5 5.25. Prove that if p is an odd prime, then the groups in Problems 5.20 and 5.21 are not isomorphic. Solution: Let G be the group defined in Problem 5.20 and let H be the group defined in Problem 5.21, for p an odd prime. Then by Problem 5.22, if g E G, gP = 1. But if G "'" H, it follows from Problem 5.24 that hp = 1 for all hE H. But by Problem 5.23, (0, l)p '7'= 1. Therefore G is not isomorphic to H. 5.26. Prove that a non-abelian group G of order p3 has a center of order p (p a prime). Solution: Let Z be the center of G. By Theorem 5.11, Z'7'= {1}. Also, Z'7'= G since G is non-abelian. then!G/Z! = p. Therefore G/Z would be cyclic and hence, by Problem 5.18, G Now if!Z! = p2, would be abelian, a contradiction. Thus!Z! = p. c. The upper central series Suppose G is a group. We shall define a series {I} = Zo c: Zl c: of subgroups Zo, Zl,... of G, called the upper central series of G. We begin by defining Zo = {I}, and ZI to be the center of G. Next we define Z2. We look at G/Zl. Since every subgroup of G/Z1 is uniquely of the form H/Z1 where H is a subgroup of G containing Zl, the center of G/Z1 is of the form Z2/Z1 (we are using Corollary 4.21, page 121). Notice that as the center of a group is a normal subgroup, Z2/Z1 is a normal subgroup of G/Z 1• There fore by Corollary 4.21, Z2 is a normal subgroup of G. In general, once Zi has been defined and proved to be a normal subgroup of G, we define Zi+dZi to be the center of G/Zi. By Corollary 4.21 it follows that Zi+1 <J G. We shall call a group G nilpotent if |
its upper central series ascends to G in a finite number of steps. Our objective in this section is to prove Theorem 5.13: A finite p-group G is nilpotent. Proof: If G = {I}, there is nothing to prove. If G =F {I}, then Zl =F {I} by Theorem 5.11. If G/Z1 is not the identity, the center of G/Zl = Zz/Zl =F ZdZl, again by Theorem 5.11. Notice that if Zl =F G, then Z2 =F Zl. Similarly if G =F Z2, Z3 =F Z2. By induction we can show that if Zi =F G, Zi+l =F Zi and thus Since G is finite, Zk = G for some k. Therefore G is nilpotent. 1 = Zo C Zl C.,. C Zi C Zi+l Problems 5.27. Prove that if a non-abelian group G is of order p3, then Z2 = G. Solution: Zl'7'= {1} by Theorem 5.11. So if Zl'7'= G, then G/Z1 is of order p or p2. Since G/Z1 is cyclic only if Zl = G (Problem 5.18), we find G/Z1 is of order p2 and hence abelian (Problem 5.19). There fore Z2/Z1 = G/Z1, i.e. Z2 = G. 5.28. Let Dn be the dihedral group of order 2n. Prove that Dn is nilpotent if n is a power of 2. Solution: Dn has the property that it contains two elements a and b such that a2 = 1, bn = 1, a-1ba = b- 1 j'= 0,1, and every element of Dn is uniquely expressible in the form aibi where..., n - 1. (See Section 3.4f, page 75, where a = l' and b = U2') i = 0,1 and Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 143 Method 1. Suppose n = 2m. Then b 2m - 1 is of order 2. Hence m-l 2m-1 2m-1 commutes |
with a; clearly b So b element of G, and so b2 2m- 1 E Zl' If m = 1, then Zl = Dn; otherwise a eo Zl' What other elements can be in the center? If aibi E Zl' then clearly a-1(aibi)a = aibi. On 1. Thus the other hand, a-1(aibi)a = aia-1bia = aib-;. Hence bi = b- i and (b i )2 = 1, i.e. bi = b2m the only possible element in the center other than b2 implies a E Zl' which is not so. Consequently Zl = {1, b2 commutes with b. Therefore b commutes with every • But as b2 E Zl' is ab 2 m-l this m-l m-l m-l }. Similarly Z2 consists of the powers of b2 Zm-l consists of the powers of b2. Now this means Thus Zm = G and G is nilpotent. m-2, •••, Zi consists of the powers of b2 • Therefore IG/Zm-11 = 4, and so G/Zm - 1 is abelian. m-'t Method 2. IDnl = some power of 2. Hence we can apply Theorem 5.13. 5.29. Prove that A4 is not nilpotent. Solution: Zl = {1} in A4, as a direct check shows. Hence Zl = Z2 =.. " and thus Zn =F A4 for every n. 5.3 DIRECT PRODUCTS AND GROUPS OF LOW ORDER a. Direct products of groups In Chapter 1 we defined the cartesian product H X K of two sets Hand K as the set of all ordered pairs (h, k), h E Hand k E K. If Hand K are groups, we can define a multi plication of elements of H X K as follows. Let (hl' k 1), (hz, k2) E H X K and define (hl' k 1) · (h2, kz) = (hlh2, klk2) (5.5) where hlh2 and klk2 are the products in the groups Hand K respectively. The multiplica tion defined in (5.5) is clearly a binary operation. The |
set H X K with binary operation (5.5) is a group. To see that H X K is a group, let (hl' k1), (h2, k2 ), (h3, k3) E H X K. Then [(hl' k 1)'(h2, k2)] • (h3, k3) (hlh2, klk2) • (h3, k3) ((hlh2)h3, (klk2)k3) (hl(h2h3), kl(k2k3)) (hl' k 1) • (h2h3, k2k3) (hl' k 1) • [(h2, k2) • (h3, k3)] Multiplication is therefore an associative binary operation on H X K. If 1 stands simulta neously for the identity element of H and of K and (h, k) E H x K, (1, l)(h, k) = (h, k) = (h, k)(l, 1), k- 1) is so that (1,1) is an identity of H x K. It is clear that if (h, k) E H x K, then (h- 1 its inverse, for (h, k)(h-l, k- 1) = (hh-l, kk- 1) = (1,1). The group H x K with binary opera tion (5.5) is called the external direct product of the groups Hand K. We often refer to H x K as just the direct product. We define the internal direct product after Proposition 5.19. If Hand K are finite groups, then it is clear that IH X KI = IHI IKI If H =1= {1} and K =1= {l} are finite groups, then H x K is neither isomorphic to H nor [H x KI =1= IKI. Therefore the direct product gives us a to K, because IH x KI =1= IHI and simple way of constructing new finite groups. For example, let C2 be the cyclic group of order 2 generated by g. C2 X C2 = {(g, 1), (g, g), (1, g), (1, 1)} 144 FINITE GROUPS [CHAP. 5 Now IC |
2 X C2 1 = 4, so we have (as we shall soon see) two non-isomorphic groups of order } where b4 = 1, and the group 4, namely: the cyclic group of order 4, C4 = {I, b, b2 C2 X C 2 •, b3 The multiplication table for C2 X C2 is (1,1) (1, g) (g,I) (g,g) (1,1) (1,1) (1, g) (g,I) (g,g) (I,g) (1, g) (1,1) (g,g) (g,I) (g,I) (g,I) (g,g) (1,1) (I,g) (g,g) (g,g) (g,I) (I,g) (1,1) Note that all the elements of C2 X C2 are of order 2. Hence C4 is not isomorphic to C2 X C2• C2 X C2 is called the Klein four group, or simply the four group. Theorem 5.14: If G = H x K is the direct product of the groups Hand K, then the sets fi K {(h,1) I hE H, 1 the identity of H} {(I, k) IkE K, 1 the identity of K} are subgroups of G. Furthermore, H "'" H, K "" K; and if a E Hand bE K, ~b = b~. Finally, G = fi K and finK = {(I, I)}, the identity subgroup of G. A A A A Proof: If (hI, 1), (h2, 1) E H, (hI, 1)(h2, 1) I = (hI,I)(h2,1) = (hlh2,1) E H, fi is clearly non-empty. Therefore fi is a subgroup of G. Similarly K then A A _ -1 -1 since h1h;1 E H. is a subgroup of G. The mapping a of H onto fi defined by is clearly an isomorphism. and b = (1, k) E K; then A A a: h --7 (h, 1), h E H Similarly K and K are isomorphic. |
Now let a = (h, 1) E H A AA AA ab = (h, 1)(1, k) = (h, k) = (1, k)(h, 1) = ba Now finK= {(1,1)}. Any element (h,k) of G can be written as (h,I)(I,k), so Gr;;,fiK. Clearly HK r;;, G. Hence G = HK and Theorem 5.14 follows. Corollary 5.15: Let G = H x K and fi,K be as in Theorem 5.14. Then every g E G A A A A can be written uniquely as a product h k where h E H, k E K. Proof: If g = (h, k), then g = (h, 1)(1, k) is an expression for g as the product of an then clearly hI = h element in H by an element in K. If we also have g = (hI, 1)(1, k1), and kl = k. Thus the expression is unique. A A A AA A A A A As a converse we have Theorem 5.16: Let G be a group with subgroups Hand K such that HnK = {I}, the elements of H commute with those of K, and HK = G. Then G "" H x K. Proof; We first show that any element g E G can be written uniquely in the form g = hk where hE Hand k E K. Since G = HK, g = hk for some hE Hand k E K. Suppose g = hlkl and g = h2k2 where hI, h2 E Hand kl' k2 E K. hlkl = h2k2 implies h;Ihl = kzk-;I. But HnK = {I}, and so h;Ihi = 1 and k2k-;1 = 1. Hence hI = h2 and kl = k2. We define the mapping a: G --7 H x K by ga = (h, k) where g = hk E G. a is a one-to one mapping, for we have shown that there is one and only one way of writing g in the form g = hk, and the elements of H x K are of the unique form (h, k). To prove a is a |
homo morphism we must demonstrate that if gl = hlkl and g2 = h2k2 are any two elements in G, then Sec. 5.3} DIRECT PRODUCTS AND GROUPS OF LOW ORDER 145 Now (hlklh2k2)a = (hlh2ktk2)a = (hlh2' klkz) = (hI' k l)(h2, k2) = (hlkl)a(h2k2)a Hence a is a homomorphism and the result follows. Note that if H,K are normal subgroups of a group G with HnK = {1}, then Hand K commute elementwise. For if h E H, k E K, h-Ik-Ihk = (h-Ik-Ih)k E K = h-I(k-Ihk) E H Therefore h-Ik-Ihk E HnK = {1}, and so Hand K commute elementwise. Clearly G = HK and Hand K commute elementwise implies Hand K are normal in G. Consequently Theorem 5.16 can be stated as follows: Corollary 5.17: Let G be a group with normal subgroups Hand K, and suppose HnK = {1}, and HK = G. Then G ~ H x K. The hypothesis of Theorem 5.16 asserts G must equal HK. But if G is a finite group It is useful to be able to and IHKI = IGI, we can conclude, since HK c:; G, that HK = G. count the number of elements in HK. We therefore prove the following proposition. Proposition 5.18: If G is a finite group with subgroups Hand K, then Proof: Let I = H n K. I is a subgroup of G and, since I c:; K, I is a subgroup of K. IHKI = IHI' IKI IHnKI Let Ikl,lk2,..., Ikn be the n distinct cosets of I in K. Thus K = Ikl U Ik2 U... U Ikn and, by Corollary 4.14, page 110, n = IKI/III = IKI/IHnKI. We claim now that HK = Hkl U Hk2 U... U Hkn For if hk E HK, then k = lkj for some |
lEI, j an integer between 1 and n. Hence hk = (hl)kj = h'kj where h' E H, as both h, l belong to H. Thus HK = HkIUHkzU·.. uHkn. Now suppose HkinHkj oF ~ for some integers i and j. Then hki = h'kj for some h, h' E H. Consequently h'-Ih = kjki-t, so kjk i- I implies that k j E Iki. Since two co sets are either equal or disjoint, Ikj = Iki. Hence ki = k j • Thus HkinHkj = ~ for i oF j and E 1= H nK. But kjk i- E I I Now IHkil = IHI, because hlki = h2ki if and only if hi = hz. Therefore IHKI = IHkl1 + IHk21 +... + IHknl since n = IK\/IHnKI. IHKI = n IHI = IHIIKI IHnKI To illustrate the use of Proposition 5.18, let G be a group of order 28 and HI and H2 subgroups of G of orders 7 and 4 respectively. HI n H 2 = {1}, because an element in HI and also in H2 must have order dividing 7 and 4. Accordingly, I HIH2 = IHI n H21 \ IHIIIH21 = 28 = \G\ 146 FINITE GROUPS [CHAP. 5 Using Proposition 5.18, we can replace Theorem 5.16 in the case of finite groups by Theorem 5.16': Let G be a finite group with normal subgroups Hand K where IHIIKI = IGI. If either (i) HnK = {1} or (ii) HK = G, then G ~ H x K. Proof: (i) HnK = {1} and [H[[KI = IG[ implies, by Proposition 5.18, [HK[ = [H[[K[/[HnK[ = [GI. Since HK (;;; G, we can conclude HK = G. But then the hypotheses of Corollary 5.17 are fulfilled and G ~ H x K. (ii) If HK = G, then IHK[ = [G[. Therefore IG[ = |
IHK[ IHIIKI IHnKI or [H n KIIGI = IHI IKI But [H[[K[ = [G[ by hypothesis. Hence HnK = {I} and, by Corollary 5.17, G ~ H x K. The concept of direct product can easily be generalized to the direct product of a finite number of groups, GI, G2, •••, Gn (n ~ 2). Let G = GI X G2 X.•. x Gn be the cartesian product of n groups. Define a multiplication in G by (gl, g2,..., gn)(g{, g~,..., g~) = (glg{, g2gf,..., gng~) for (gl, g2,..., gn), (g{, g;,...'g~) E G. G is then a group (see Problem 5.30 below) called the (external) direct product of the groups GI, G2,..., Gn • We denote G by n IT Gi • i=l In Chapter 6 we will define the direct product of an infinite number of groups dif ferently. Proposition 5.19 below and Corollary 5.15 will provide a link between the two definitions. Proposition 5.19: Let Hand K be subgroups of a group G. If (i) hk = kh for all h E Hand k E K and (ii) every element g EGis a unique product of an element in H and an element in K, (i.e. g = hk, hE H, k E K; and if g = hlkl' hI E H, kl E K, then h = hI and k = leI), then G ~ HxK. Proof: We need only prove· HnK = {1} to fulfill the hypotheses of Theorem 5.16. Suppose g E HnK. Then g = h·1 = 1· k for some hE Hand k E K. But condition \ii) implies h=1 and k=1. Therefore g=l and HnK={1}. If G is a group with subgroups Hand K satisfying conditions (i) and (ii), G is said to be the internal direct product of Hand K, and we write G = H Q9 K. By Proposition 5.19, HQ9K~H |
xK. Problems 5.30. Let G = G I X G2 X •.. X Gn be the cartesian product of n groups. Define a multiplication in G by (gl,g2'··.,gn)(g;,g~,.·.,g~) = (glg;,g2g~,·· ·,gng~) for (gl' g2,..., gn), (g;, g~,..., g~) E G. Show that G is a group. Solution: The multiplication is clearly an associative binary operation in G, since multiplication is associai = 1,2,..., n, then (1,1,...,1) is the inverse tive in each Gi• If 1 stands simultaneously for the identity of Gi, is clearly the identity of G. of g in G. If g = (gl,g2,...,gn) E G, then (g~l,g;I,...,g;:l) 5.31. If H ~ Hand K ~ K, where H, H, K and K are groups, then H X K ~ Ii X K. Solution: If a: H.... Ii and fJ: K.... K are isomorphisms, we define y: H X K.... H X K by y: (h, k).... (ha, kf3), hE H, k E K. y is a one-to-one mapping, for (ha, kf3) = (h'a, k'f3) if and only if ha = h'a Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 147 and k(3 = k' (3. Since a and (3 are one-to-one mappings, ha = h' a and k(3 = k' (3 h = h' and k = k'. To show y is a homomorphism, let (h, k), (h', k') E H X K. Then if and only if [(h, k) • (h', k')]y (hh', kk')y = «hh')a, (kk')(3) = (hah' a, k(3k' (3) (ha, k(3)· (h'a, k'(3) = (h, k)y(h', |
k')y Finally, it is clear that y is an onto mapping. 5.32. Show that G = H X K is an abelian group if and only if Hand K are both abelian groups. Solution: Suppose Hand K are abelian groups. Letting (h, k), (h', k') E G, (h, k) • (h', k') = (hh', kk') = (h'h, k'k) = (hi, k') • (h, k) and so G is abelian. Conversely, suppose G is abelian. Let h, h' E H. Then if 1 is the identity of K, (h, l)(h', 1) = (h', l)(h, 1) or (hh',l) = (h'h,l). But this implies hh' = h'h. Hence H is abelian. Similarly we can show K is abelian. 5.33. Consider the groups C2 X K 4, C4 X C2, Cs where Cn is the cyclic group of order nand K4 the four group, i.e. the non-cyclic group of order 4, described above. Are any two of these groups isomorphic? Is anyone non-abelian? Solution: Let C2 = gp(a), C4 = gp(b), and Cs = gp(g). We look at the set of elements of order 2 in each group. Since every isomorphism maps elements of order 2 onto elements of order 2, if there are more elements of order 2 in one group than in another, these groups cannot be isomorphic. Every element (oF 1) of C2 X K4 is of order 2, for (c, k)2 = (c2, k 2) = (1,1) and (1, k)2 = (1, k 2) = (1,1) for any k E K 4 • Cs on the other hand has only one element of order 2, namely g4, because (gi)2 oF 1 if i oF 4, 0 :<C i:<c 7. Now C4 X C2 has at least one element of order 4, (b,l), and at least two elements (b 2, a) and (1, a). Therefore no two of the groups are isomorphic. As |
C 2, C 4, K4 and Cs of order 2, are abelian, Problem 5.32 implies C2 X K 4, C4 X C2 and Cs are also abelian. Thus we have exhibited three non-isomorphic abelian groups of order 8. 5.34. If Cn and Cm are the cyclic groups of order nand m respectively and Cn X Cm =' Cnm, the cyclic group of order nm. (n, m) = 1, then Solution: Say Cn = gp(g) and Cm = gp(h). Consider the order of the element (g, h) in Cn X Cm. We claim that the order of (g, h) is nm. If (g, h)k = (1,1) for some k, then (gk, hk) = (1,1) and so gk = 1 and hk = 1. Since the order of g is n and the order of h is m, m I k and n I k. Hence k is divisible by nm. On the other hand, (g, h)nm = (gnm, hnm) = 1 and so the order is nm. Accordingly, Cn X Cm = gp«g, h)). Therefore Cn X Cm =' Cnm' since all cyclic groups of the same order are isomorphic (Theorem 4.7, page 103). 5,35. Show that Cs2 is not isomorphic to Cs X Cs (where Cn is the cyclic group of order n), for any integer 8 > 1. Solution: Since Cs2 is a cyclic group it has, by Theorem 4.9, page 105, one and only one subgroup of order 8. But Cs X Cs has two subgroups of order 8, namely gp«l, g)) and gp«g, 1)), where g is the generator of Cs' Since subgroups of a given order are mapped onto subgroups of the same order by any iso morphism, Cs2 cannot be isomorphic to Cs X Cs. 5.36. Show that for any prime p there are exactly two non-isomorphic groups of order p2. Solution: By Problem 5.19, page 140, we know that any group of order p2 is abelian. Cp 2, the cycl |
ic group of order p2, and Cp X Cp, where Cp is the cyclic group of order p, are two non-isomorphic groups of order p2 (Problem 5.35). To see that these are the only possible groups of order p2, consider a 148 FINITE GROUPS [CHAP. 5 subgroup H of order p in G, a group of order p2. Such a subgroup exists by Corollary 5.12. H is cyclic, since p is a prime. Let a (;l H. The order of a is either p or p2. If the order of a is p2, G is a cyclic group generated by a. If!gp(a)! = p, then gp(a)nH = {I}, for b (# 1) E gp(a)nH implies H = gp(b) and gp(a) = gp(b), since a group of prime order has no proper subgroups. Also!gp(a)!!H! = p2 and, as G is abelian (Problem 5.19), gp(a) <J G and H <J G. Therefore, using Theorem 5.16', we conclude G "'" gp(a) X H. But gp(a) X H "'" Cp X Cp by Problem 5.31. Hence G "'" Cp X Cpo 5.37. Show (Hl X H 2) X H3 "'" Hl X H2 X H 3. Solution: Define 'lr: (H l X H 2) X H3 ~ Hl X H2 X H3 by 'lr: ((hl' h 2), h3) ~ (hv h2' h3) for hl E H, h2 E H2 and h3 E H 3. Clearly 'lr is an onto mapping. If ((hl' h 2), h3)'lr = ((hl' h z), h 3)'lr, then (hv h2' ha) = (hv h2' h3) and consequently hl = hl' h2 = h2 and ha = ha• Therefore 'lr is one-to-one. To show 'lr is a homomorphism, let ((hl' hz), h3) and ((hl, h 2), ha) E (Hl X Hz) X H 3. Then [(( |
h v h 2), ha)((hl' h 2), ha»)'lr = ((hl' h2)(hl' h 2), h3h3)'lr = ((hlh l, h 2h 2), h3ha)'lr and so 'lr is an isomorphism. (!tlhl' h2h2' h3h a) = (hv h2' ha)(hl' h2' h3) ((hv h 2), h3)'lr((hl' h 2), ha)'lr h. Groups of small order: orders p and 2p As an application of the Sylow theorems and the theorems of Section 5.3a we will find, up to isomorphism, all groups of order less than 16. We will use Cn to denote the cyclic group of order n, and K4 to denote the four group. Recall that K4 is defined to be C2 x C2 • We refer to the notation of Section 5.3a. Let us put 1 = (1,1), x = (1, g), y = (g, 1) and z = (g, g). The multiplication table for K4 is We note that xy = z, xz = y and yz = x. Notice that the multiplication table is sym metric in x, y and z. If we put x = a and y = b, then z = ab and we can write the multi plication table in the form 1 a b ab 1 a b 1 a b a 1 ab ab ab b b ab ab 1 a b a 1 There is, up to isomorphism, clearly only one group of order 1. If p is a prime, any group of order p is cyclic (Problem 4.48, page 110). Up to isomor phism, there is one and only one cyclic group of order p (see Theorem 4.7, page 103). Thus In particular, the only groups of there is one and only one group of order p, p a prime. order 2,3,5,7,11 and 13 are cyclic. There are precisely two non-isomorphic groups of order 4, namely C4 and K4 (Section 5.3a and Problem 5.36). Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 149 Next we show there are precisely two groups in each case of order 6, 10 or 14 |
. Note that 6 = 2·3, 10 = 2·5 and 14 = 2·7, so these groups are of order 2p for some prime p 7'= 2. Let G be a group of order 2p, p an odd prime. By Problem 5.10, G has exactly one subgroup K of order p, and either or (i) (i) exactly one subgroup H of order 2 (ii) precisely p subgroups of order 2. In this case the group G is a cyclic group of order 2p. To see this notice that H is a unique Sylow 2-group and K is a unique Sylow p-subgroup; so, by Problem 5.7, page 133, H <J G and K <J G. Furthermore HnK = {I}, for any element common to H and K must have order dividing 2 and p and hence is the identity. Clearly JHJJKJ = 2p = JGJ. Therefore by Theorem 5.16', G "'" H x K. But Hand K are cyclic groups of order 2 and p respectively. Thus by Problem 5.34, G is cyclic of order 2p. (ii) Let K = gp(a) wher.e aP = 1. Since K is the only subgroup of order p, b (/:. K implies b2 = 1. Clearly, G = K u bK. Hence G consists of the distinct elements 1, a, a2 Now if i = 0,1,...,p-l, then, •••, aP-l, b, ba, ba2, •••, baP- 1 (bai)2 1 and bai = aP-ib (5.6) (5.7) since bai (/:. K and each element of G outside K is of order 2. Also (bai)2 = (bai)(bai) = 1 implies bai = (bai)-l = (ai)-lb- 1 = aP-ib Now if G is any group of order 2p, then it is either of type (i) or (ii). If G is of type (i), then by our analysis it must be a cyclic group of order 2p. By Theorem 4.7, page 103, cyclic groups of the same order are isomorphic. Hence all groups of order 2p having property ( |
i) are isomorphic. Suppose G is of type (ii). Then, arguing as above, G has a subgroup K = gp(a) of order p and an element b of order 2 such that G = {I, a,..., aP-l, b, ba,..., bap- l } where for i = 0,1,...,p-l, (bai)2 = 1 and bai = aP-ib (5.8) The mapping a: G ~ G defined by a: ai ~ ai, a: b ~ b, a: bai ~ bai (i any integer) is an isomorphism. First, a is a mapping; for if ai = ai, p divides i - j and hence ai = ai. Consequently a is well defined on ai. If bai = bai, then ai = ai and so p divides i - j and ai = ai. Hence a is well defined on the bai. As a: ai ~ ai, bai ~ bai (i = 0,1,..., p - 1), a is one-to-one and onto. a is also a homomorphism, for gl, g2 E G implies gl = biai and g2 = bSat for some choice of j, s E {O, I} and i, t E {O, 1,2,..., p -I}. Using equations (5.7) and (5.8), we obtain, when s = 0, (glg2)a = [(biai)(at)]a = (biai+t)a = biaHt = biaiat = (biai)a(at)a = glag2a and when s = 1, (glg2)a = [(biai)(bat)]a = (bibap-iat)a = (bHlaP-i+t)a bHlaP-i+t = biaibat = (biai)a(bat)a = glag2/X Therefore any two groups of order 2p of type (ii) are isomorphic. 150 FINITE GROUPS [CHAP. 5 So far we have shown that up to isomorphism there are at most two possible groups of order 2p, p a prime. This does not mean that there |
exist two non-isomorphic groups of order 2p, for each prime p. But from Theorem 4.7, page 103, we know that for each positive integer n there exists a cyclic group of order n, and from Section 3.4f, page 75, we know that for each n the order of the dihedral group D" is 2n and Dn is not cyclic for n > 2 (it is not even abelian). There are therefore, up to isomorphism, exactly two groups of order 2p for each prime p cI= 2: one a cyclic group and the other Dp. In particular there are exactly two non-isomorphic groups of order i, i = 6,10,14. It is worthwhile summarizing our method of finding all groups of order 2p. We first showed that if a group had order 2p it had to be isomorphic to one of two possible groups. The isomorphism in case (ii) was shown by using the fact that the elements of such a group (As those equations determine the group up to had to satisfy equations (5.7) and (5.8). isomorphism, they are usually called defining relations for the groups; they will be discussed in detail in Chapter 8.) After demonstrating the isomorphism, we proved that each of the possible groups exists by exhibiting a group of each type. c. Groups of small order: orders 8 and 9 Let G be a group of order 8. There are at least three non-isomorphic abelian groups of order 8: Cs, C2 X C4 and Cz x K4 (see Problem 5.33). We show that if G is abelian it is isomorphic to one of these three groups. If G has an element of order 8, G is cyclic. If G has no element of order 8 but has an element a of order 4, let H = gp(a). Let bEG - H. If b is of order 4, b2 is an element of order 2 and lies in H (since the coset decomposition of G is Hu bH). As a2 is the only element of H of order 2, b2 = a2. Hence (ab)2 = a2b2 = a2a2 = 1. Since ab ~ H, we may assume that there exists an element x E G - H of order 2 (x = b if b is of order |
2 or else x = ab). Let X = gp(x). XnH = {I}. Therefore by Theorem 5.16', G "'" X X H. Since X "'" C2 and H "'" C4, we conclude G "'" C2 X C4 • If G has no elements of order 4 or 8, all its non-identity elements are of order 2. Let a, b be distinct elements of order 2 in G. Let A = gp(a), B = gp(b). Then AB is a group by Problem 4.62, page 114. Now AnB = {I} and IAIIBI = IABI. SO, by Theorem 5.16', AB "'" A X B and consequently AB"", C2 X C2 • Let c E G - AB and C = gp(c); then CnAB = {I}. Thus G "'" (C 2 X C2) X C2 = K4 X Cz. We conclude that there are, up to isomorphism, exactly three abelian groups of order 8. Assume G is a non-abelian group of order 8. G has an element of order 4 and no elements of order 8; for if g EGis of order 8, G "'" Cs. On the other hand if all elements of G are of order 2, then (ab)2 = 1 for any a, bEG and consequently ba = a2bab2 = a(abab)b = ab contrary to our assumption that G is non-abelian. Let a E G be an element of order 4, and put H = gp(a). Then G = HuHb for some bEG. Also H <l G, as it is of index 2 (Problem 4.69, page 116). b2 E H; for if not, the co sets H, Hb, Hb 2 would be distinct, and this would contradict [G: H] = 2. We have four possibilities for b2 : (i) bZ = a, (ii) bZ = aZ (iii) b2 = a3, or (iv) b2 = 1., If (i) or (iii) occurs, clearly G = gp(b), contrary to our assumption. Thus (ii) and (iv) are the only possibilities. (ii) b2 = a2 • Since H <l G, b-1 |
ab E H. As a is of order 4, so is b-1ab. Thus b-1ab = a or a3 • If b-1ab = a, then ab = ba. But every element of G can be written as aib or ai for some integer i, since G = HuHb. Hence ab = ba implies G is abelian, contrary to our assumption. Thus b-1ab = a3 or ab = ba3 (5.9) Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 151 Since G = HUHb, the elements of G can be expressed as 1, a, a2 If a group of this type actually exists, then equation (5.9), b2 = a2 and a4 = 1 provide us with enough information to construct its mUltiplication table., b, ab, a2b, a3 b., a3 1 1 a a2 a3 b 1 a b ab b b ab ab a2b a3b ab a2b a3b b a a a2 a3 1 a2 a3 1 a a3 1 a a2b a3b a2 a3b a3b a2b ab ab b a3b a2b a2b ab b a3b a3b a2b ab b a2 a3 1 a b ab 1 a a2 a3 ab a2b a3 1 a a2 b a a2 a3 1 To calculate the products in the table, we used the fact that ab = ba3 and b2 = a2 imply ba = a3b since a3 b = a2(ba3 ) = b3a3 = b(a2a3 ) = ba. Table 5.1 If G is another non-abelian group of order 8 with an element a of order 4 and an element b (l gp(a) such that b2 = a2, then as in our argument above, G = {i, a, a2, a3, b, a2b, iib, ab 3 } with the elements satisfying the equations a4 = 1, a2 = b2, ab = ba3 from which we find a multiplication table for G which is identical to Table 5.1 except that a is substituted for a and b for b. The mapping a: G ~ G defined by a: a ~ a and |
a: aib ~ aib, i = 0,1,2,3, is clearly an isomorphism. Table 5.1 also shows that a group of order 8 of this type actually does exist, for the table defines a group. To see this, notice that the product of any two elements is again an element, i.e. the table defines a binary operation, 1 is an identity element, and every element has an inverse. The only difficulty is checking that the binary operation is associative. This involves much calculation (the reader should check some of the calculations himself). We shall give another description of this group in Problem 5.40. This group is 'called the Quaternion group and has the interesting property that all its subgroups are normal and yet it itself is not abelian (Problem 5.43). We now move on to a discussion of (iv). (iv) b2 = 1. Let K = gp(b); then HnK = {I} and G = HK. Now H <J G so that • As in (ii), b-1ab = a implies b-1ab E H and, since a is of order 4, we have b-1ab = a or a3 G is abelian. Hence b-1ab = a3, which leads to The elements 1, a, a2 b2 = 1 and a4 = 1 enable us to construct the following multiplication table., a3, b, ab, a2b, a3b are the distinct elements of G. Equation (5.10), ba = a3b (5.10) 152 FINITE GROUPS [CHAP. 5 b b ab ab a2b a3b ab a2b a3b b a2b a3b b ab a3 1 a a a a2 a3 1 a2 a3 1 a 1 1 a a2 a3 b 1 a b a2 a3b a3b a2b ab b a3 1 a a2 ab a2b a2 a3 1 a a a2 a3 1 1 a a2 a3 ab ab b a3b a2b a2b ab b a3b a3b a2b ab b Table!:.2 As in part (ii), any non-abelian group of order 8 having property (iv) is isomorphic to G. Such a group exists, for Table 5.2 |
also defines a group. This group is isomorphic to the dihedral group D 4, the group of symmetries of a square (see Problem 5.38). The two groups given in Tables 5.1 and 5.2 are not isomorphic because the group of Table 5.1 has exactly one element a2 of order 2, whereas the group of Table 5.2 has five, b, ab, a2b and a3b. Thus we have shown that there are exactly two elements of order 2: a2 non-isomorphic non-abelian groups of order 8. To summarize, there are five non-isomorphic groups of order 8, three abelian and two. non-abelian. Since 9 = 32, we know by Problem 5.36 that there are two and only two non-isomorphic groups of order 9, namely C9 and C3 X C3 • d. Groups of small order: orders 12 and 15 To complete our list of all groups up to order 15, we must find all possible groups of order 12 and 15. Because 12 = 3.22, we know that a group G of order 12 has at least one Sylow 2-subgroup of order 22 and at least one Sylow 3-subgroup of order 3. The third Sylow theorem tells us that the number of Sylow 2-subgroups is congruent to one modulo 2 (i.e. S2 = 1 +2k for some integer k) and S2 divides IGI. When k = 0, S2 = 1; and when k = 1, S2 = 3. If k > 1 it is clear that 1 + 2k does not divide 12. We therefore have two possibilities: G has exactly one Sylow 2-subgroup or G has exactly three Sylow 2-subgroups. A similar argument shows that G has exactly one Sylow 3-subgroup or G has exactly four Sylow 3-subgroups. Therefore we have four possibilities: (i) S2 = 1 and S3 = 1 (ii) S2 = 1 and S3 = 4 (iii) S2 = 3 and S3 = 1 (iv) S2 = 3 and S3 = 4 Notice that because the Sylow 2-subgroup has order 4 it must be isomorphic to C4 or K 4, and the Sylow 3....gubgroup must be isomorphic to C3 (Section 5.3b). We treat |
each case separately. Sec.5.3J DIRECT PRODUCTS AND GROUPS OF LOW ORDER 153 (i) Let F be the Sylow 2-subgroup and T the Sylow 3-subgroup of G. Then F <J G and T <J G, since a Sylow p-subgroup is a normal subgroup if it is unique (Problem 5.7, page 133). Furthermore, Fn T = {I} since any element in the intersection must have order dividing 3 and 4 and so must be the identity. Moreover, WIITI = 12. Hence by Theorem 5.16', G 25' F x T. We have two possibilities for F: (a) F 25' C4 or (b) F 25' K 4 • Thus G 25' C4 X C3 25' Cl2 (by Problem 5.34) or G 25' K4 x C3; both these groups are abelian. These are the only abelian groups of order 12, for if G is abelian any two conjugate subgroups are equal. Hence by Theorem 5.3, page 131, 82 = 83 = 1. In cases (ii) through (iv) we assume G is a non-abelian group. Furthermore, let F be any Sylow 2-subgroup of G and T any Sylow 3-subgroup of G. Then Fn T = {I} and, by IFTI = IFllT[llFn TI = IFIITI = IGI. Thus G = FT. If It = tl Proposition 5.18, page 145, for all IE F and t E T, then G is abelian since gl, g2 E G implies gl = Iltl and g2 = ht2 for some!t.12 E F and t l, t2 E T. Now as F and T are both abelian groups, glg2 = Ittd2t2 = 12tdltl = g2g1 Hence we also assume if F is any Sylow 2-subgroup of G and T any Sylow 3-subgroup of G, that it is not the case that It = tl for all IE F and t E T. (ii) 82 = 1 and 83 = 4. Let F be the (unique) Sylow 2-subgroup and T be a Sylow 3-sub group. There are two possibilities: (a |
) F 25' C4 and (b) F 25' K 4 • We treat each case separately. (a) Let F = {I, a, a2, a3} where a4 = 1, and T = {I, b, b2} where b3 = 1. im plies F <J G. Thus b-Iab E F. If b-Iab = a, then ab = ba. But this implies every element of F commutes with every element of T, contrary to our assumption. Therefore since a has order 4, b-Iab # a2 and b-Iab # 1 so that b-Iab = a3 or ab = ba3. We show that under these assumptions gp(ba) = G. (ba)2 = b(ab)a = b2a4 = b2. So (ba)3 = (ba)2ba = b2ba = a. Hence gp(ba) contains a and b and thus coincides with G. Then G is cyclic, contrary to assumption. There is therefore no non-abelian group of order 12 with 82 = 1, 83 = 4 and Sylow 2-subgroup iso morphic to C4 • 82 = 1 (b) Let F = {I, x, y, z} be the four group as given in Section 5.3b, and T = {I, c, c2 } where c3 = 1. As in part (a), F <J G so that c-Ilc E F for all IE F. Now by assumption c-Ilc # I for at least one I E F. Suppose c-Ixc # x (the other cases are similar). We may assume c-Ixc = y. is argued similarly.) Let us, as in Section 5.3b, put x = a, y = band z = abo Then ac = cb, implies c-Ibc = cbc- 1 which implies a = cbc- 1• Now c-Ibc # a, for c-Ibc = a or b = &bc-2 • Then, as & = c- l and c- 2 = c, b = c-Ibc. Hence a = b, a contra diction. Similarly c-Ibc # band c-Ibc # 1. Then c-Ibc = ab and c-l(ab)c = (c-Iac)(c |
-Ibc) = bab = b2a = a. Consequently.the equations (The other case, c-Ixc = Z, ac = cb, bc = cab, :..abc = ca, a2 = b2 = 1, c3 = 1, ab = ba j\.., ;-JJ- (5.11) hold in G. Now 1, c, c2 determine distinct cosets of F in G. Therefore the elements of G are ~. S z <I (;,. S;f/ 6- Using equations (5.11), we can write down the multiplication table for G, as shown in Table 5.3 below. 154 FINITE GROUPS [CHAP. 5 1 1 c c2 a b ab ca cb 1 c a b ab ca cb c C c2 1 c2 1 c a a b b ab ab ca ca cb cb cab cab ca cb cab c2a c2b c2ab c2a c2 b c2ab a b c ca cab ab ca c cb b a 1 cab cb c cb c2ab 1 ab cab c2a ab ca c2b c2b ab b c 1 a a b cb c2 cab cb c2ab c2 c2a c2ab cab c ca c2b c2a c2 cab cab c2a ca c c2 c2ab c2b c2a b cab c2ab c2b ab c2b ab ca c2ab c2 c2a c2ab a cb c2b c2a c2 b 1 Table 5.3 1 a ab a 1 b c2a c2b c2ab a ca b cb ab cab c2b c2a c2 c2 c2ab c2b c2ab b 1 ab cb c c2 a ab 1 ca cab cab c c2a 1 b a c cb ca By a similar argument to that used in the discussion of the non-abelian groups of order 8, any group of order 12 with 82 = 1 and 83 = 3 is isomorphic to G. Moreover, Table 5.3 defines a group: the table defines a binary operation, the identity is 1, and every element clearly has an inverse. The associativity of the operation must also be checked, an even more tedious task than in the case of a group of order 8. The |
alter nating group A4 is a group of this type (Problem 5.38). (iii) 82 = 3 and 83 = 1. Let F be a Sylow 2-subgroup and T = {1, e, e2} (e3 = 1) be the (a) F = {1, a, a2, a3} "'" C4 and Sylow 3-subgroup. Again we have two possibilities: (b) F = {l,x,y,z} "'" K 4 • (a) T, being a unique Sylow 3-subgroup, is normal in G and so a-lea E T. We may assume a-lea""" e, otherwise the group is abelian. Hence a-lea = e2 and ea = ae2. Also, e2a = eae2 = ae4 = ae. The equations.. which determine a multiplication table for this group a r e i \ t.,.vL (5.12) The distinct elements of G are then and we obtain Table 5.4 below. We conclude, by an argument similar to that used in the discussion of the non abelian groups of order 8, that any group of order 12 with 82 = 3, 83 = 1 and in which the Sylow 2-subgroups are cyclic of order 4, is isomorphic to the group G defined in the table. Again it can be checked that the table defines a group, so that a group of this type exists. We have as yet not encountered an example of this type of group, but in Problem 5.41 we show that there is a group of 2 by 2 matrices which is isomorphic to G. Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 155 a2 a3e a3c2 a2e a a a2 a3 1 a2 a3 1 a 0.,3 1 a 1 a e 1 1 a a2 a3 e e2 ae2 a2e a3e2 ae a2e2 a3e e2 1 ae ae a2e2 a3e e2 ae2 a2e a3e2 e ae2 a2e2 a3e e2 ae a2e2 a3e2 ae2 ae2 a2e a3e2 a2e2 a3e e2 e |
ae a3e2 e ae2 a2e a a2 a3 e e ae ae e2 a2e a3e ae2 a2e2 a3e2 ae ac2 a2e a3e c a2c2 a3c2 e2 a2e a2e2 a3e ae a3e2 e2 ae2 e ac e a a2e2 a3 ae2 a2e2 a2 a3e c2 ae a 1 e a a2 a3 ae2 a2 a3e2 a2e a2 a3 1 a3e2 e2 ae2 1 a a2 e2 ae a2e2 a3 a2e a3e a3e2 e2 Table 5.4 1 a ae2 a2c2 a2e a3 a3e e a2 a3 1 1 a a3e e ae a3 e ae a2e 1 a a2 (b) F = {1, x, y, z} and T = {1, C, c2 }. Since T <J G, we have I-lcl E T for all IE F. By assumption, for at least one IE P, I-lcl # c. We may therefore assume, with out loss of generality, that X-ICX = c2• Again, as in Section 5.3b, put x = a, y = b and z = abo Then ca = ac2 ) = (ca)c 2 = ac 2c2 = ac,. 2 I.e. c a - ac. • Note that c2a = c(ca) = c(ac 2, a, ca, c2a} We claim that 8 = {1, C, c2 181 = 6 for cia = ci is a subgroup. We leave to the reader the task of checking that the product of any two elements in 8 is again in 8 (ca = ac 2 and c2a = ac make this task easy). The identity 1 is in 8, and on inspec tion we find every element in 8 has an inverse in 8: c- l = c2, (C2)-1 = c, a-I = a, (ca)-l = ca, (c 2a)-1 = c2a. Hence 8 is a |
subgroup of G. (i = 1 or 2; and j = 0, 1 or 2) implies a E T, a contradiction; cia = a (i = 1 or 2) implies ci = 1, a contradiction; and ca = c2a implies c = 1, a contradiction. 8 is is not equal to ca) and hence is isomorphic to D3 since non-abelian (ac = c2a there is only one non-abelian group of order 6 (up to isomorphism). Now [G: 8] = 2 (by Problem 4.69, page 116). Hence b-lcb E 8. As b-lcb is an element of order 3, it is either c or c2, as all other elements of 8 are of order 2. We shall now choose an element hE F, h 6!: 8, such that h-lch = c. If b-lcb = c, let h = b. If on the other hand b-lcb = c2 let h = abo Recall that a-lca = c2 • Then (ab)-lc(ab) = b-l(a-lca)b = b- I C2b = b-lcb· b-lcb = • c2 = c. Hence there exists an element h 6!: 8 in F (Le. b or ab) such that h -lch = c. c2 Consider H = gp(h). Clearly 8nH = {1}, 8 and H commute elementwise, and 1811HI = IGI, and so G ~ 8 X H by Theorem 5.16. But 8 ~ D3 and H ~ C2. We therefore conclude that any group G with 82 = 3, 83 = 1 and Sylow 2-subgroup isomorphic to K 4, is isomorphic to D3 X C2• The dihedral group De is a group of this type (Problem 5.38). implies 8 <J G, (iv) 82 = 3, 83 = 4. Since distinct cyclic groups of order 3 intersect in the identity element, the four Sylow 3-subgroups have together 9 distinct elements. A Sylow 2-subgroup is of order 4. Since the intersection of a group of order 4 and a group of order 3 can only be the identity, it follows that the number of distinct elements in the 4 3-Sylow subgroups and a single |
2-Sylow subgroup is 12. But IGI = 12, so there cannot be another distinct Sylow 2-subgroup. Thus there is no group of type (iv). 156 FINITE GROUPS [CHAP. 5 To summarize, we have shown that there are up to isomorphism exactly three non-abelian groups of order 12. They are the groups (ii) (b) with S2 = 1. S3 = 4 and the Sylow 2-subgroup ~ K 4 ; see Table 5.3. (Such a group is isomorphic to A4.) (iii) (a) with S2 = 3, S3 = 1 and the Sylow 2-subgroup ~ C4; see Table 5.4. is isomorphic to a group of 2 by 2 matrices given in Problem 5.41.) (Such a group (b) with S2 = 3, S3 = 1 and the Sylow 2-subgroup ~ K 4 • (Such a group is isomorphic to D3 X C2 which is isomorphic to Ds. See Problem 5.38.) Clearly no two of these groups are isomorphic. The abelian groups of order 12 are K4 X C3 and C12 • Thus including the abelian groups, there are five non-isomorphic groups of order 12. If G is of order 15, we have seen that G is cyclic (Section 5.1a). The following table gives the number of non-isomorphic groups of order 1 through 15. Order of group No. of groups 10 11 12 13 14 15 2 1 5 1 2 1 The reader will agree that finding all groups of a given order is difficult. Indeed there is not even a general method of determining how many non-isomorphic groups of a given order there can be. Problems 5.38. Show that (i) The dihedral group D4 is a group of order 8 isomorphic to the group given in Table 5.2, page 152. (ii) The alternating group A4 is isomorphic to the group given in Table 5.3. (iii) The dihedral group Ds is isomorphic to D3 X C2• Solution: (i) D4 is a non-abelian group of order 8 and as such is isomorphic to one of the groups given in Section 5.3c, Tables 5.1 and 5.2. A check of Table 5.1 shows that |
there is only one element of order 2, namely a2. But D4 is the symmetry group of the square (Section 3.4f, page '75). A reflection T is of order 2 and if u is a rotation of 90°, TU is of order 2 as can be seen in the discussion of these groups in Chapter 3. Therefore D4 must be isomorphic to the group given in Table 5.2. (ii) A4 is a non-abelian group of order 12. Hence it is either isomorphic to the group of Table 5.3 or 5.4 or to D3 X C2 (see page 155). As can be seen from the multiplication table for A4 given in Chapter 3, page 63, A4 has exactly three elements of order 2, namely U2' U5 and U8. Now the group of Table 5.4 has only one element a2 of order 2, so that it could not be isomorphic to A 4• We have shown in Problem 5.1, page 131, that A4 has no subgroup of order 6 and D3 is iso morphic to a subgroup of D3 X C2 by Theorem 5.14. Hence A4 is not isomorphic to D3 X Cz. Thus it must be isomorphic to the group given in Table 5.3. (iii) Ds is of order 12, and is not abelian. Ds is the symmetry group of the hexagon and therefore has a subgroup of order 6, namely the rotation of 60° about the center. So it is not the case that Ds ~ A 4, and hence Ds cannot be isomorphic to the group of Table 5.3. Also, a reflection followed by a rotation is an element of order 2. Since there are six such elements in D s, it cannot be isomorphic to the group of Table 5.4 as this group has only one element of order 2. The only other possibility is that Ds ~ D3 X C2• 5.39. Find a cyclic subgroup of order 6 in D3 X C2• Solution: Let a E D3 be of order 3 and b (# 1) E C2• Consider the element (a2,1), (a, b)3 = (1, b), (a, b)4 = (a,l), (a, b)5 = (a2, b), and (a, |
b)6 = (1,1). Hence gp«a, b» subgroup of D3 X C2 of order 6. (a, b) E D3 X C2• (a, b)2 = is a cyclic Sec. 5.3] DIRECT PRODUCTS AND GROUPS OF LOW ORDER 157 5.40. and B = -1 0 where i = V-i. A and B have nonzero Consider the matrices A = determinants and thus are elements in the group of all 2 X 2 matrices with nonzero determinants. Show that gp(A, B) is a group of order 8 which is isomorphic to the quaternion group (Table 5.1, page 151). (0 i) i 0 ( 0 1) Solution: By direct calculation we find o -1 A2 = (-1 0) (-1 0) o -1 AB = - t ( ~ -i) -i 0) 0 0 i ( A4 (~ n I, (identity matrix) A3B C 0) o -i A2B = G -~) Let G = {I,A2,A3,B,AB,A2B,A3B}. We claim that G = {l,A,A2,A3,B,AB,A2B,A3B} gp(A, B). Clearly, G c; gp(A, B). To show G = gp(A, B), we need only show G is a group, as A, BEG. Note G is a subset of the group of 2 X 2 matrices with nonzero determinant. Hence the elements of G satisfy the associative law. By direct calculation we can show that G is closed under matrix multiplication; the equation BA = A3B simplifies the calculations, e.g., and A3B = BA Furthermore every element of G has an inverse in G, e.g. using B3A = AB we have (A3B)-1 = B-lA -3 = B3A = AB Checking all these details enables us to conclude that G is a group of order 8 and G = gp(A,B). G is non-abelian, since AB #- BA, so G is either isomorphic to the group of Table 5.1 or Table 5.2. Because G has only one element of order 2, it cannot be isomorphic to the group of |
Table 5.2. Thus G is isomorphic to the quaternion group of Table 5.1. 5.41. Consider the matrices A = (~ ~) and B = (~ ;), where i = A and < is a nonreal com plex cube root of 3 (so, in particular,.3 = 1 and < #- 1). A and B have nonzero determinants and thus are elements in the group of all 2 X 2 matrices with nonzero determinants (see Section 3.5b, page 81). Show that gp(A,B) is a group of order 12 which is isomorphic to the group given in Table 5.4, page 155. Solution: We find by direct calculation that B3 B2 A3 A2B A2B2 (~ ~) A2 = (-1 0) o -1 ( ~ -i) - t 0 A4 = G ~) Let H = {A,A2,A3,A4,B,B2,AB,A2B,A3B,AB2,A2B2,A3B2}. We claim H = gp(A,B). Clearly H c; gp(A, B). To prove H = gp(A, B) we need only show H is a group, as A, BE H. Note that H is a subset of the 2 X 2 matrices with nonzero determinant. Hence the elements of H satisfy the associative law. To check that H is closed under matrix multiplication, first note that A -lBA = B2. Then, for example, (-< 0) o _<2 (-~ -i~2) AB2 = C~2 ~) (~ ~) AB = (~ i~2) (-~ 0) o -< ( 0 -U) -w 0 A3B2 A3B A2A· (A -lBA)A2B = A3B2A2B = A4(A -lBA)(A -lBA)AB = B2B2AB = BAB = AA -lBAB = AB2B = A Also the inverse of, for example, (A3B) is given by (A3B)-l = B2A = AA -lB2A = A(A -lBA)(A -lBA) AB E H 158 FINITE GROUPS [CHAP. 5 Checking all these details |
enables us to conclude that H is a group of order 12 and H = gp(A, B). The mapping a: a ~ A, and c ~ B is an isomorphism of the group of Table 5.4 and H. This obtains because H satisfies the equations BA = AB2, B2A = AB, B3 = J, A4 = J (J the identity matrix). These are the exact counterparts of equations (5.12), page 154. Con sequently the multiplication table for H is obtained from that for the group of Table 5.4 by renaming via a. 5.42. Show that a group G of order 48 has a normal subgroup =F {1} or G. (Very difficult.) Solution: By the first Sylow theorem, G has a Sylow 2-subgroup of order 16. By the third Sylow theorem, 82 = 1 + k2 for some integer k and 821 48. The only odd divisor of 48 is 3, hence 82 = 1 or 3. If 82 = 1, then the Sylow 2-subgroup is unique and therefore normal (Problem 5.7, page 133). Suppose 82 = 3. Let Hand K be two of the Sylow 2-subgroups. As HnK is a proper subgroup of H, IHKI = IHnKII16. Then IHnKI = 8; for if IHnKI "" 4, IHIIKI/IHnKI ~ 16 X 16/4 = 64, which contradicts our assumption that IGI = 48. Since both H and K are of order 16, HnK, as a subgroup of index 2, is normal in both Hand K (Problem 4.69, page 116). Hence Ht;;;,NG(HnK) and Kt;;;,NG(HnK). Letting N = NG(HnK), we have HKt;;;,N. Thus INI ~ IHKI = IHIIKI/IHnKI = 32. As INI divides 48 and INI ~ 32, INI = 48 and so N = G. Because a group is normal in its normalizer, we have HnK <J G. then, by Proposition 5.18, page 145, 5.43. Show that all subgroups of the quaternion group G are normal subgroups. page 151.) (G is given in Table 5 |
.1, Solution: It is sufficient to check that the cycli,c groups are normal in G. For if S is any subgroup, 8 E S, and x E G, then X- 18X E gp(8) implies X- 18X E S. Using the multiplication table we can check that for x = a or b and any 8 E S, x- 1SX E gp(s). (We leave this check to the reader.) This is sufficient to prove the result, for every element of G is of the form aib or ai for i = 0,1,2,3. 5.4 SOLVABLE GROUPS a. Definition of solvable groups To introduce our concepts we will begin with an example. If P is a group of order pT where p is a prime, then we showed that P has a series of subgroups Pi with where each Pi <J P i+ 1 and (5.4), page 139). {1} = Po k P 1 k... k P T = P (5.13) [PH 1: Pi] = p for each integer i = 0,..., r - 1 (see equation Let G be a group and suppose it has a series of subgroups {1} = Go k G1 k... k GT = G If each Gi <J GH1 for i = 1,..., r -1, then (5.14) is called a subnormal series of (for) G. With this definition, (5.13) is a subnormal series of P. (5.14) If (5.14) is a subnormal series for G and [Gi +1: Gi] is some prime (dependent on i), for i=0,...,r-1, G is called a solvable gTOUp and (5.14) is called a solvable series for G. Accordingly we conclude that P is solvable and that (5.13) is a solvable series for P. If (5.14) is a subnormal series for G and Gi+t!Gi is simple, i.e. Gi+t!Gi has no normal sub groups other than GHt!Gi and the identity, for i = 0,...,r-1, then (5.14) is said to be a composition series for G. To see that (5.13) is a composition series for P, note that Pi |
+t!Pi is a cyclic group of order p and hence is simple. We call the factor groups G;/Gi+1 of the subnormal series (5.14) the factors of (5.14). Sec. 5.4] SOLVABLE GROUPS 159 We shall discuss composition series in greater detail in Section 5.5. We remark that not all groups have a composition series but finite groups do (Section 5.5a). Our main con cern in this section is the concept of solvable group. Historically, solvable groups arose in the attempt to find a formula for the roots of an nth degree polynomial f(x) = anxn + an _Ix n - 1 +... + a1X + ao in terms of the coefficients ai. The formula sought was one which involved the coefficients ao,..., an of the polynomial, integers, and the operations addition, subtraction, multiplica tion and division, and a finite number of extraction of roots. For example, if n = 2, then x = obtained by such a formula, we say f(x) = 0 is solvable by radicals. is a formula giving the roots of (5.15). If the roots of f(x) can be -a1 ± ya2 - 4a2aO 2 1 a2 (5.15) From the Fundamental Theorem of Algebra we know that an nth degree polynomial with complex coefficients has n complex roots. Let F be the "smallest" field (see Section 3.6b, page 86, for a definition of field) of complex numbers containing the coefficients 0,; of f(x). By saying F is the smallest field we mean that if H is a field containing the coefficients 0,;, then Fe H. Let E be the smallest field containing F and the roots of f(x). Now the set of automorphisms of E forms a group under the composition of mappings (see Theorem 3.15, page 87). The automorphisms of E which map every element f E F onto itself is a subgroup G of the group of all automorphisms of E. The group G is called the Galois group of the polynomial f(x). In the beginning of the 19th century, the French mathematician E. Galois proved (essentially) the following extraordinary theorem: An equation f(x) = 0 is solvable by radicals |
if and only if the Galois group of f(x) is solvable. It turns out that not all equations of degree ~ 5 are solvable by radicals because the symmetric group Sn (For details see Birkhoff and MacLane, A Survey of Modern is not solvable for n ~ 5. In Section 5.5e we will prove that Sn is not solvable. Algebra, Macmillan, 1953.) Problems 5.44. Show that the symmetric group 8 n is solvable for n = 1,2,3. Solution: 8 1 = {,}; then 8 1 has the solvable series {,} C;;;81 and is therefore solvable. 8 2 = {G!), G ~)} then {,} C;;;82 is a solvable series for 8 2 and so 8 2 is solvable. Let 8 3 = {" (11) (12, 7"1, 7"2' 7"3} where 2 G 2!) G 2 ~) 3 (11 (12 7"1 7"2 2 1 G 2!) G 2 ~) 1 [83 : H] = 2. Hence by Problem 4.69, page [83 : H] = 2, G 2 ~) G 2!) 3 7"3 is a cyclic subgroup of 8 3, Also, is a solvable series for 8 3, since [H: {,}] = 3 and Now H = {" (11) (12} 116, H <l 8 3, Thus {,} C;;; H C;;; 8 3 and so 8 3 is solvable. 5.45. Show that 8 4 is solvable. Solution: The alternating group A4 is a subgroup of order 12 in 8 4, Then [84 : A4] = 2 and A4 <l 8 4 by Problem 4.69, page 116. We have seen in Problem 5.1, page 131, that A4 has no subgroup of order 6. Now A4 is a group with a unique Sylow 2-subgroup F of order 4 and F ~ K 4, the four group (see Problem 5.38(ii), page 156, and Section 5.3d, page 153). Since F is a unique Sylow 2-subgroup, F <l A4 and [A4: F] = 3. F, being a four group, has a normal subgroup K of order 2 |
. Accordingly, 160 FINITE GROUPS [CHAP. 5 is a subnormal series for S4. As solvable group. {l} eKe F c A4 C S4 [K: {l}] = 2, [F: K] = 2, [A4: F] = 3 and [S4: A4] = 2, S4 is a b. Properties of, and alternative definition for, solvable groups An important property of solvable groups is given in Theorem 5.20: If G is a finite group and N <J G is such that N and GIN are solvable groups, then G is also solvable. Proof: Let {N} <: HI <: H2 <:... <: Hk = GIN be a subnormal series for GIN with [Hi+l: Hi] = Pi, Pi a prime. By the correspondence theorem (Theorem 4.19, page 120, and Corollaries 4.20 and 4.21), there are subgroups Hi in G such that Hi <l Hi+l, HJN = iii and [Hi+ I: Hi] = [Hi+ 1: iii] = Pi (i = 0,1,..., k - 1). Therefore (5.16) is a series of subgroups of G with Hi <l Hi+! and [Hi+l: Hi] = Pi. Now N is also a solvable group. Hence N has a series {1} = Ko <: KI <: Kz <:... <: K! = N where [Ki+l : Ki] is a prime number (i = 1,2,..., l-1). Putting the series (5.16) and (5.17) together, we obtain (5.17) {1} = Ko <: KI <:... <: K! <: HI <: H2 <:... <: Hk = G which is a solvable series for G. The proof is complete. Note. In contrast to Theorem 5.20, it is not always true that a group G has a property if both a normal subgroup N and GIN have the property; for example, the four group K4 has a normal cyclic subgroup N of order 2 which is cyclic and K41N is cyclic, but K4 itself is not cyclic. Corollary 5.21: |
If G is a finite abelian group, G is solvable. Proof: We use induction on the order n of G. If JGJ = 2, the result holds trivially. Assume that any abelian group of order less then n is solvable. Suppose P J n for some prime p. Then by Proposition 5.9, page 137, G has an element of order p. Let a be such an element. If P =/= n, then Jgp(a)J < JGJ so that gp(a) is solvable by the induction assumption. Furthermore, since G is abelian, gp(a) is a normal subgroup of G and JGIgp(a)J < JGJ. Hence GIgp(a) is solvable by our induction assumption, and so, by Theorem 5.20, G is solvable. If p = n, then {1} <: G is a solvable series and G is therefore solvable in this case too. The following theorem leads to an alternative definition of solvability. 'I'heorem 5.22: G is a solvable group if and only if G is finite and has a subnormal series {l} = Ko <: KI <:... <: Kn = G (5.18) where Ki+t/Ki is abelian (i = 0,1,...,n-1). Proof: Let G be a solvable group. Then G has a subnormal series with factors of prime order and hence cyclic. Since a cyclic group is abelian, the solvable series of G is a series of type (5.18). Conversely, assume G has a subnormal series (5.18) with K+t/Ki abelian. We prove that G is solvable by induction on n, the length of the subnormal series If n = 1, then G is abelian since G =' Kt/Ko which is abelian by assumption. (5.18). Hence by Corollary 5.21, G is solvable. Assume that any finite group which has a subnormal series of length less than n in which the factor groups of consecutive terms of the series are abelian, is solvable. Let G have subnormal series (5.18) of length n. Then K n - I has a subnormal series of length n - 1 |
, namely Sec. 5.4] SOLVABLE GROUPS 161 Ko C; K1 C;... C; K n - 2 C; K n - 1 with Ki+dKi abelian for i = 0,1,..., n - 2. Hence by the induction assumption, K n- 1 is solvable. But GIKn- 1 is abelian and hence solvable. Using Theorem 5.20, we conclude G is solvable. In the theory of infinite groups one usually defines a group G to be solvable if it has a subnormal series (5.18) with K+dKi an abelian group (i = 0,1,..., n -1). By Theorem 5.22 this is equivalent to our original definition for finite groups. Since this formulation of solvability is more general, we shall henceforth use it as our definition of solvability. Note that the infinite cyclic group is an example of a group that does not fit the old definition but does fit the new. Using this new definition we prove Theorem 5.23: Let G be a solvable group. Then (i) any subgroup of G is solvable and (ii) if N <J G then GIN is solvable. Proof: (i) Let {1} = Ho C; H1 C;..• C; Hn = G be a subnormal series of G with Hi+dHi abelian for i = 0,1,..., n - 1. We show that if K is a subgroup of G, {1} = (K n Ho) C; (K n H 1) C;... C; (K n Hn) = K (5.19) is a subnormal series with KnHi+dKnHi abelian. First we notice that KnHi = (KnHi+1)nHi (i = 0,1,..., n - 1) and that KnHn = K. Now Hi <J Hi+1, and K n H i + 1 is a subgroup of H i + 1. Applying the subgroup isomorphism theorem (Theorem 4.23, page 125) inside the group Hi+1 with subgroup KnHi+1 and normal subgroup Hi, we obtain and Since (KnHi+1)nHi = KnHi, it follows that KnHi <J KnHi+1 and |
(K n Hi+1)/(K n Hi) "" ((K n Hi+1)Hi)IHi But (K n Hi+ l)Hi C; Hi +1, so that we have (K n Hi+1)Hi Hi Hi+1 C; Hi Now Hi+dHi is abelian by assumption and hence so is (KnHi+1)H;/Hi • Therefore (5.19) is a subnormal series for K with abelian factors K n Hi + 1 and consequently K is solvable. K n Hi (ii) Let G have subnormal series {1} = Ho C; H1 C; H2 C; •.. C; Hn = G where Hi+dHi is abelian. Now N <J G. Consider the natural homomorphism v: G ~ GIN Any subgroup of G is mapped by v onto a subgroup of GIN. In particular let Hi = Hiv. We assert Hi <J i!;n But this follows from Problem 4.82, page 122 (with 0 = VIHi+1 and H i + 1 = G). Next we assert that Hi+dHi is abelian. Let x = Xv, ii = yv (x,y E H i +1) be two elements of Hi+1. Then since Hi+dHi is abelian, xy = yxd where dE Hi. Thus (xy)v = (xv)(yv) = (yv)(xv)(dv) 162 FINITE GROUPS [CHAP. 5 But dv E iii. Consequently Xiiiyiii = Xyiii = (xy) Viii Therefore iii+diii is abelian. Thus (yv )(xv )dviii (YV)(XV)Hi {N} = iio <J iil <J... <J iin = GIN is a subnormal series of GIN with abelian factors, and so GIN is solvable. Now that we have a definition of solvable group that applies to infinite groups, we will extend Theorem 5.20 to infinite groups. Theorem 5.24: If N <J G and GIN and N are solvable groups, then so is G. Proof: Let {N} = Ho C iil C H2 C... C iik = GIN be a subnormal series for G |
IN in which Hi+ d Hi is abelian, i = 0,..., k - 1. By Corollary 4.21, page 121, there are subgroups Hi in G such that Hi <J H i +l and H.!/N ==- iii (i = 0,..., k). By the factor of a factor theorem (Theorem 4.22, page 121), Hi+dHi = (Hi+dN)/(HdN) 2" Hi+dHi• Hence the factors Hi+dHi are abelian. Also, N has a series {1} = Ko C Kl C K2 C... C KI = N with Ki+dKi abelian for i= 0,1,...,l-1. Therefore {1} = Ko C Kl C '" C KI = N = Ho C Hl C... C Hk G is a subnormal series whose factors are abelian. Thus G is solvable. Problems 5.46. Show that all groups G of order p2, pq or p2q, where p and q are distinct primes, are solvable. (Hard.) Solution: If IGI = p2, then G is abelian (Problem 5.19, page 140) ann. G is solvable. If jGI = pq then from Problem 5.8, page 133, if p < q, G has one and only one subgroup H IG/HI = p, hence G/H is abelian. As H is of of order q. By Problem 5.7, page 133, H <J G. Now order q, it is abelian. Tberefore we have the subnormal series {I} c; H c; G with abelian factors and so G is solvable. If IGI = p2q then 8p = 1 + kp divides p2q, and so the prime factors of 1 + kp must be p or q. Clearly p does not divide 1 + kp. Therefore 1 + kp = 1 or q. If 1 + kp = 1, then the Sylow p-sub group H is normal in G (Problem 5.7). As H is of order p2, H is abelian (Problem 5.19). Thus we have a subnormal series {I} c; H c; G with G/H abelian |
(IG/HI = q) and H/{I} abelian. Hence G is solvable. Suppose, however, that 1 + kp = q; then q > p. Let K be a Sylow q-subgroup of G. The num ber of such Sylow q-subgroups is 1 + lq. Again 1 + lq is not divisible by q, and so 1 + lq = 1, p or p2. But as q > p, the only possibilities are 1 + lq = 1 or p2. Case (i): 1 + lq = 1. In this case there is only one Sylow q-subgroup K and (by Problem 5.7) IG/KI = p2. Hence K is abelian and G/K is abelian (Problem 5.19), and it IKI = q and K <J G. follows that G is solvable. Case (ii): 1 + lq = p2. We will show that this case does not arise by showing that G would contain too many elements. We have assumed that G has q Sylow p-subgroups (of order p2) and p2 Sylow q-subgroups (of order q). Any two distinct subgroups of order q intersect in the identity, so there are p2(q - 1) = p2q - p2 distinct elements of order q in G. Also, G has at least 2 Sylow p-subgroups and hence there are at least p2 distinct elements in G of order p or p2. In the above calculations we have not counted the identity, so in all G has at least p2q - p2 + p2 + 1 = p2q + 1 elements, which is absurd, and we conclude that case (ii) does not arise. Sec. 5.5] COMPOSITION SERIES AND SIMPLE GROUPS 163 5.47. Show that every nilpotent group is solvable. Solution: Consider the upper central series of G, a given nilpotent group. Zi+ 1 is defined by the fact that Zi+ llZi is the center of GIZi and of course this implies Zi + llZi is abelian. Hence G is solvable. {l} = Zo k Zl k Z2 k... k Zn = G 5.48. Prove that the converse of Problem 5.47 is false, i |
.e. not all solvable groups are nilpotent. Solution: The symmetric group 8 3 is solvable (Problem 5.44). A check of the multiplication table for 8 3 on page 57 shows that the center of 8 3 is just the identity {t}. But this implies that the upper central series for G never ascends to G. Thus 8 3 is not nilpotent. 5.49. Let G be any group and let G(i) be defined for all positive integers i by GO) = G', the derived group of G, and G(i+ 1) = (G(i))'. Prove that G is solvable if and only if G(n) = {l} for some integer n. Solution: Let G(n) = {l}. Then {l} = G(n) k... k GO) k G is a subnormal series for G and G(i) IG(i+ 1) is abelian. Hence G is solvable. Now let G be solvable. Then there exists a subnormal series {l} = Hr k... k Ho = G with H/Hi+l abelian. By Problem 4.68, page 116, HJHi+ 1 abelian implies H i+ l:! H;. We prove, by induction on i, that Hi:!GW, i=1,2,...,r. For i=l this is true since Hl:!H~=G'=GO). Suppose our assertion is true for i=n, i.e. Hn:!G(n). Then H~:!(G(n»)'=G(n+1). But Hn+l:!H~, so In particular then, {l} = Hr:!G(r). Accordingly, H n+l :!G(n+1). Therefore Hi:!GW for all i. G(r) = {l} and the result follows. 5.5 COMPOSITION SERIES AND SIMPLE GROUPS a. The Jordan-Holder Theorem In Section 5.4a we introduced the idea of a composition series for a finite group G. We recall that a series of subgroups of G {I} = Go ~ Gl ~... ~ Gk = G (5.20) is a composition series for G if Gi <J Gi+ 1 for i = 0 |
,1,..., k - 1 and Gi + dG i is simple, i.e. has precisely two different normal subgroups. This latter statement carries with it the implication that Gi 0/= Gi + 1 for i = 0,..., k -l. We observe first that every finite group G has a composition series. The easiest way to If IGI = 1, then G has precisely one see this is by induction on the order, IGI, of G. composition series: {I} = Go = G Suppose then that IGI 0/= 1 and that every group of order less than IGI has a composition series. Now if G is simple, then {I} = Go ~ G l = G is the only composition series for G. If G is not simple, let N be a normal subgroup of G, N 0/= {I}, N 0/= G. We may suppose that N is the largest normal subgroup of G, that is, if M <J G and M 0/= G, then IMI ~ INI. By induction, N has a composition series {I} = No ~ Nl ~... ~ N z = N 164 FINITE GROUPS [CHAP. I) We claim that {I} = No C Nt C... C Nl C G is a composition series for G and note that to prove this we need only show that GINl is simple. But if GINl is not simple, it has a non-trivial normal subgroup. By the corollary to the correspondence theorem this subgroup is of the form KINl where K is a normal subgroup of G. But as K:::J N l, this means that IKI > INri which contradicts the choice of N l • Therefore every finite group has a composition series. This proof does not suggest that if a group has two composition series then they are related. Surprisingly they are. In order to explain this relationship we associate with the composition series (5.20) two notions. First we term k the length of the series. Second we call the factor groups Gi+t!G; the composition factors of the series (5.20). The relation ship between composition series is given by Theorem 5.25 (Jordan-Holder): Every finite group G has at least one composition series. The lengths of all composition series for G are equal. Finally if {I} G and {I} Ho |
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